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### Math book

1. 1. Hi,This book is based on some quicker methodsfor different numerical problems, hence useful for BankExams.MULTIPLICATION TABELE (10x10)1 2 3 4 5 6 7 8 9 101 1 2 3 4 5 6 7 8 9 102 4 6 8 10 12 14 16 18 203 9 12 15 18 21 24 27 304 16 20 24 28 32 36 405 25 30 35 40 45 506 ThinkAs8*44 *8,..36 42 48 54 607 49 56 63 708 68 72 809 81 9010 100MULTIPLICATION TABLE (25x10)21 21 42 63 84 105 126 147 168 189 21022 22 44 66 88 110 132 154 176 198 22023 23 46 69 92 115 138 161 184 207 23024 24 48 72 96 120 144 168 192 216 24025 25 50 75 100 125 150 175 200 225 2501 2 3 4 5 6 7 8 9 1011 11 22 33 44 55 66 77 88 99 11012 12 24 36 48 60 72 84 96 108 12013 13 26 39 52 65 78 91 104 117 13014 14 28 42 56 70 84 98 112 126 14015 15 30 45 60 75 90 105 120 135 15016 16 32 48 64 80 96 112 128 144 16017 17 34 51 68 85 102 119 136 153 17018 18 36 54 72 90 108 126 144 162 18019 19 38 57 76 95 114 133 152 171 19020 20 40 60 80 100 120 140 160 180 200MULTIPLICATION TABLE (11-20 x11-20)11 12 13 14 15 16 17 18 19 20
2. 2. 11 121 132 143 154 165 176 187 198 209 22012 144 156 168 180 192 204 216 228 24013 169 182 195 208 221 234 247 26014 196 210 224 238 252 266 28015 225 240 255 270 285 30016 256 272 288 304 32017 289 306 323 34018 324 342 36019 361 38020 400For quickest math-1. Many diagrams are given in almost every example which would be the reminder of theprocess when ucompletely learn the techniques.2. Try to understand the colors, every color is explaining the steps and other important things.Edited byRAKES PRASADFirst we learn how to find square, square root and cube roots with multiplication techniques.Rule no 1. You are to remember squares of 1 to 32. They are given as followsNo Square No Sq No Sq No Sqno sq1 1 9 81 17 289 25 625 33 10892 4 10 100 18 324 26 676 34 11563 9 11 121 19 361 27 729 35 12254 16 12 144 20 400 28 784 36 12965 25 13 169 21 441 29 841 38 14446 36 14 196 22 484 30 900 41 16817 49 15 225 23 529 31 9618 64 16 256 24 576 32 1024N.B. The square of 38 contains 444 and the sq of an odd prime no 41 contains two perfectsquares as 42= 16 and 92= 81.Rule No 2. Squares of numbers ending in 5 :we are showing by givinganexample.For the number 25, the last digit is 5 and the previous digit is 2. to multiply the previous digit 2by onemore , that is, by 3. It becomes the L.H.S of the result, that is, 2 X 3 = 6. The R.H.S of the resultis
3. 3. 52, that is, 25. Thus 252= 2 X 3 / 25 = 625. In the same way, 1052= 10 X 11/25 = 11025; 1352= 13 X14/25 = 18225; see the figure below…….Now we are extending the formula asRule no 3: If sum of the last two digits give 10 then you can use the formula butL.H.S shouldbe same. Check the examplesEx 1 : 47 X 43 .See the end digits sum 7 + 3 = 10 ; then 47 x 43 = ( 4+ 1 ) x 4 / 7 x 3= 20 / 21 = 2021.Example 2: 62 x 682 + 8 = 10, L.H.S. portion remains the same i.e.,, 6.Next of 6 gives 7 and 62 x 68 = ( 6 x 7 ) / ( 2 x 8 ) = 42 / 16 =4216.Example 3: 127 x 123127 x 123 = 12 x 13 / 7 x 3 = 156 / 21 = 15621. (see the fig )Example 4. 39523952= 395 x 395 = 39 x 40 / 5 x 5 = 1560 / 25 = 156025.You can rem the formula as 1252=(122+12)25=(144+12)25=15625. Bothare same things but it can use in bigger case easily.Rule 4: Also we can extend the formula where the sum of last two digitsbeing 5 asEx 1. 82 x 83=(82+8/2) /3 x 2=(64+4) / 06=6806 i.e.( n2+n/2) n is even but note thatthe R.H.S term should be of two digits as 06 .( in this ex.)
4. 4. Ex 2.181 x 184=(182+9) / 04=(324+9) / 04=33304.But we know what are u thinking of? Yes, if the number be odd then whathappen?And ur ans is as follows:Ex 3: 91 x 94= (92+9/2) / 04=(81+4) / 54. Surprised? We just pass ½ as 5 in the nextrow since ½ belongs to hundred’s place so 1/2 x 100=50. Just pass 5 in suchcases.See the fig.Ex 4 . 51 x 54= (25+2) / 54 =2754 Ex 5. 171 x 174= (289+8) /54=29754Rule no 5 . Now we are discussing the general method of finding squares ofanynumber. See the chart first.See 52=25And 102= 100 , so on.Now u can understand the tricks.base range Trick alternative50 26-74 25(±)( read as 25 plusminus)100 76-126 100(±)2(±) Given no(±)150 126-174 225(±)3(±)200 176-224 400(±)4(±)250 226-274 625(±)5(±)300 276-224 900(±)6(±)350 326-374 1225(±)7(±)400 376-424 1600(±)8(±)450 426-474 2025(±)9(±)500 476-524 2500(±)10(±)……… ………….. ……………………….1000 976-1024 100,00(±)20(±)1500 1476-1524 225,00,00(±)30(±)2000 1976-2024 4,000,000(±)40(±)For base 50: Now we are explaining one by one through some examples.Ex 1. 432
5. 5. step 1. Find base –no as 50-43=07Step 2. Sq the no as 072=49.write it on the right side.Step 3. Subtract the no from 25 as 25-7=18.write it on extreme left.Step 4. Now this is Ur ans. 1849Ex 2. 572step 1. 57-50=7Step2. 72=49Step3. 25+7=32step4. Ans is 3249.Ex 3. 692step1. 69-60=19 , 192=361Step2. carry over 3 and write 61 in right side.Step3 . 25+19=44, now add 44 with 3 (carry number) and writeit with extreme left.i.e. (25+19+3)Step 4. ur ans is 4761Ex4 . 38 2 step1. 50-38=12 , 122=144 ,carry 1 and write 44 at right side .Step2. now (25-12)+1=14 i.e. Carry number always adds (itnever subtract). So ur ans is 1444.So, 1. At first see the given no is it more or less than Ur base. If it is morethen addition rule and if it be less, the n subtraction rule will beapplied. That’s why (±) sign is given2. And always add the carried no.3. range is chosen making (±) 24 from base .4. Trick is done by squaring the base excluding a 0 (zero) such for base 50 :5 2 =25.for base 350 , 35 2 = 12255. for(±) divide the base without zero by 5 as for base 100: 10/5=2,so trickis100 (±)2(±) and for base 350: 35/5=7 ,and the trick is 1225(±)7(±) andso on.
6. 6. Now we are giving more examples to clear the fact, a regular practice can makeumaster of the art.ok, seeThe following examples.For base 100Ex 5. 892Step1. 100-89=11 ,112=121 ,carry 1Step2. 100-(2 x 11 )+1=79, so the ans is 7921Also u can use the alternative formula here,”given no (±)” such asStep 1. 100-89=11,sq is 121,carry 1Step2. 89-11+1=79 ,ass is 7921(I think this process is better for numbers whose base is near 100)Ex 6. 117²Step1. 117-17=17, 172=289 ,carry 2Step2. 117+17=134, 134+2=136 ,ans is 13689For base 150Ex 6. 139²1. 150-139=11 , sq is 121 ,carry 12. 225-3×11=225-33=192, 192+1=193 ,ans is 19, 321Ex 7. 164 ² 1. 164-150=14, 14 ² =196, carry 12. 225+3(14)=225+42=267 ,267+1=268 ,ans is 26,896Ex 8. 512 ² 1. 12 ² =144 ,carry 1, 2500+10(12)+1=2621, ans is 262144Try yourself to various numbers to learn the technique quickly. now we learntofind the square root .before starting note that…Rule no 6:unit digit of perfectsquare1 4569unit digit of square root 1,9(1+9)=102,82+8=1055+5=104,64+6=10
7. 7. 3,73+7=10Notes : 1 . any no end with 2,3,7,8 can’t be a perfect square.2. any no ends with odd no zero’s can’t be a perfect square.Now check the given boxes…range sq range sq range sq range1²=1 1-3 9²=81 81-99 17²=289 289-323 25²=625 625-6752²=4 4-8 10²=100 100-120 18²=324 324-360 26²=676 676-7283²=9 9-15 11²=121 121-143 19²=361 361-399 27²=729 729-7834²=16 16-24 12²=144 144-168 20²=400 400-440 28²=784 784-8405²=25 25-35 13²=169 169-195 21²=441 441-483 29²=841 841-8996²=36 36-48 14²=196 196-224 22²=484 484-528 30²=900 900-9607²=49 49-63 15²=225 225-255 23²=529 529-575 31²=961 961-10238²=64 64-80 16²=256 256-288 24²=576 576-624 32²=1024 1024-1088Rule: 1. separate the two numbers of extreme right.2. find the range of the square of the rest, write it on left side.3. Multiply the range with its preceding number and see that the separatednumber is more or less than the multiplied number.4. if more, then choose the large no of unit digit as the R.H.S.Ex1. √ (2601)1. 26 / 012. 26 falls on the range of 5 ,write 5 in the left.3. now 5×6=30 ,26 is smaller than 30.4. so choose 1 as the right side no. ans is 51Ex2. √ (6241)1. 62 / 412. 62 falls in range of 7 ,write 7 in left.3. 7×8=56 i.e.,62 is more than 56,so we choose 9 as the rightsideEx3. √ 27041. 27 / 042. 27 falls in range of 5, so 5×6=30 ,but 27 is less than 30, soWe choose 2 between 2 & 8 (allocated for 4)3.so the ans is 52But it is easy if the number ends with 5 such asEx 4. √ (99225)1. 992 /252. Here u can insert 5 as right side because there is only 5 allocatedfor 5 and no choice need here.
8. 8. 3. clearly 992 falls in the range of 31 and the ans is 315Exercise 1. √ 34225 2. √ 105625 3. √ (0.00126025) 4. √ 22095. √ ___________________2916 6. √ 2116 7. √ 15129 8. √ 161299. √ 55696 10. √ 66564 11. √ 8.8804 12. √ 0.0010112413. √ 0.125316Rule no 7: let us know how to find cube root quickly. See the box first.unit digit ofperfect cube1 2 3 4 5 6 7 8 9unit digit ofcube root1 82+8=1073+7=104 5 6 37+3=1028+2=109n.b. 1, 4 , 5, 6, 9 has no change as they were in the sq formula and 2,3,7,8has therecompliment with 10, i.e., 2+8=10, 3+7=10 etc.Ex1. (46656)1/31. 46 / 6562. write 6 for 6 on right side3. see 46 falls in the range of 3 34.so the ans is 36cube range cube range cube range1 =1 1-7 6 =216 216-342 11 =1331 1331-17272 =8 8-26 7 =343 343-511 12 =1728 1728-21963 =27 27-63 8 =512 512-728 13 =2197 2197-27434 =64 64-124 9 =729 729-999 14 =2744 2744-33745 =125 125-215 10 =1000 1000-1330 15 =3375 3375-409521=926125=15625Ex2 . cube of 0.00200483831. 0.0020048 / 3832. write 7 for 3 , and see 2048 falls in the range of 12
9. 9. 3. so the ns be 0 .123n.b. (se 3 places to shift 1 decemal place. )Rule no 8: Let us know to find the cube of any two digits number.Ex 1. (18) 3Step1. Find the ratio of the numbers such as 1:8 in this example.Step 2. Write cube of first digit and then write three successive terms inhorizontal line which are multiplied by the ratio. in this ex asStep 3. Make double of second and third terms and write them just belowtheir own positionsStep 4. Add successive terms (carry over if more than one digit) and u willget the required result .Lets see the methods …1. 18 , that is 1:84 24 512. 1 3 =1 / 1x8=8 / 8 x 8=64 / 64 x 8 =5123. 8 x 2=16 64 x 2=1284. ------------------------------------------------------------( 4 +1) ( 24 +8+16) ( 51 +64+128) 51 25 = 4 8 = 24 35. So the ans is 5832.Ex 2. (33)3Rakes Prasad , 2008It contains some vedic methds for multiplication of two,threeandmore digits,finding h.c.f. ofr two polynomial equations.Edited byRakes Prasad, 2008Rule no (1) If R.H.S. contains less number of digits than the number ofzeros in the base, the remaining digits are filled up by giving zero or zeroeson theleft side of the R.Note: If the number of digits are more than the number of zeroes in hebase, the excess digit or digits are to be added to L.H.S of theanswer.Case 1: Let N1 and N2 be two numbers near to a given base inpowers of 10, and D1 and D2 are their respectivedeviations(difference)from the base. Then N1 X N2 can be represented asEx. 1: Find 97 X 94. Here base is 100.
10. 10. .Ex. 2: 98 X 97 Base is 100.ans: 95/06=9506Ex. 3: 75X95. Base is 100. .Ex. 4: 986 X 989. Base is 1000Ex. 5: 994X988. Base is 1000 .Ex. 6: 750X995Case 2:Ex. 7: 13X12. Base is 10.Ex. 8: 18X14. Base is 10Ex. 9: 104X102. Base is 100.104 04102 02¯¯¯¯¯¯¯¯¯¯¯¯Ans is 106 / 4x2 = 10608 ( rule - f ) .Ex. 10: 1275X1004. Base is 1000.1275 2751004 004...............................................1279 /275x4 =1279 / 1 / 100 (rule f) =1280100Case ( iii ): One number is more and the other is less than the base.Ex.11: 13X7. Base is 10Ex. 12: 108 X 94. Base is 100.Ex. 13: 998 X 1025. Base is 1000.Find the following products by the formula.1) 7 X 4 2) 93 X 85 3) 875 X 9944) 1234 X 1002 5) 1003 X 997 6) 11112 X 99987) 1234 X 1002 8) 118 X 105Rules no 2,3,4 of chapter no 1 can also be extended asEg. 1: consider 292 x 208. Here 92 + 08 = 100, L.H.S portionis same i.e. 2292 x 208 = ( 2 x 3 ) / 92 x 8 / =736 ( for 100 raise theL.H.S. product by 0 ) = 60736.Eg. 2: 848 X 852Here 48 + 52 = 100, L.H.S portion is 8 and its next number is 9.848 x 852 = 8 x 9 / 48 x 52720 = 2496= 722496.[Since L.H.S product is to be multiplied by 10 and 2 to be carriedover as the base is 100].Eg. 3: 693 x 607693 x 607 = 6 x 7 / 93 x 7 = 420 / 651 = 420651.Find the following products .
11. 11. 1. 318 x 312 2. 425 x 475 3. 796 x 7444. 902 x 998 5. 397 x 393 6. 551 x 549(2) One less than the previous1) The use of this sutra in case of multiplication by 9,99,999.. is asfollowsa) The left hand side digit (digits) is ( are) obtained by deduction 1fromthe left side digit (digits) .b) The right hand side digit is the complement or difference betweenthemultiplier and the left hand side digit (digits)c) The two numbers give the answerExample 1: 8 x 9Step ( a ) gives 8 – 1 = 7 ( L.H.S. Digit )Step ( b ) gives 9 – 7 = 2 ( R.H.S. Digit )Step ( c ) gives the answer 72Example 2: 15 x 99Step ( a ) : 15 – 1 = 14 Step ( b ) : 99 – 14 = 85 ( or 100 – 15 )Step ( c ) : 15 x 99 = 1485Example 3: 24 x 99 Answer :Example 5: 878 x 9999 Answer :find out the products64 x 99 723 x 999 3251 x 999943 x 999 256 x 9999 1857 x 99999(3) Multiplication of two 2 digit numbers.Ex.1: Find the product 14 X 12The symbols are operated from right to left .Step i) : Step ii) :Step iii) :Ex.4: 32 X 24Step (i) : 2 X 4 = 8Step (ii) : 3 X 4 = 12; 2 X 2 = 4; 12 + 4 = 16.Here 6 is to be retained. 1 is to be carried out to left side.Step (iii) : 3 X 2 = 6. Now the carried over digit 1 of 16 is to beadded. i.e., 6 + 1 = 7. Thus 32 X 24 = 768Note that the carried over digit from the result (3X4) + (2X2)= 12+4 = 16 i.e., 1 is placed under the previous digit 3 X 2 = 6 andadded.After sufficient practice, you feel no necessity of writing in this wayandsimply operate or perform mentally.(4) Consider the multiplication of two 3 digit numbers.
13. 13. in order, we get the answer ( 82 + 1 ) / ( 08 – 03 ) / ( 100 – 16 )= 83 / 05 / 84 = 830584 . 16 becomes 84 after taking1 from middlemost portion i.e. 100. (100-16=84). _ Now 08 - 01 = 07 remains in themiddle portion, and 2 or 2 carried to it makes the middleas 07 - 02 = 05. Thus we get the above result.Eg.(3): 9983 Base = 1000; initial deficit = - 2.9983 = (998 – 2 x 2) / (- 6 x – 2) / (- 2)3 = 994 / 012 / -008= 994 / 011 / 1000 - 008 = 994 / 011 / 992 = 994011992.Find the cubes of the following numbers using yavadunam sutra.1. 105 2. 114 3. 1003 4. 10007 5. 926. 96 7. 993 8. 9991 9. 1000008 10. 999992.(6) Highest common factor:Example 1: Find the H.C.F. of x2 + 5x + 4 and x2 + 7x + 6.1. Factorization method:x2 + 5x + 4 = (x + 4) (x + 1)x2 + 7x + 6 = (x + 6) (x + 1)H.C.F. is ( x + 1 ).2. Continuous division process.x2 + 5x + 4 ) x2 + 7x + 6 ( 1x2 + 5x + 4___________2x + 2 ) x2 + 5x + 4 ( ½xx2 + x__________4x + 4 ) 2x + 2 ( ½2x + 2______0Thus 4x + 4 i.e., ( x + 1 ) is H.C.F.OUR PROCESSi.e.,, (x + 1) is H.C.FExample 2: Find H.C.F. of 2x2 – x – 3 and 2x2 + x – 6Example 3: x3 – 7x – 6 and x3 + 8x2 + 17x + 10.Example 4: x3 + 6x2 + 5x – 12 and x3 + 8x2 + 19x + 12.(or)Example 5: 2x3 + x2 – 9 and x4 + 2x2 + 9Add: (2x3 + x2 – 9) + (x4 + 2x2 + 9) = x4 + 2x3 + 3x2.÷ x2 gives x2 + 2x + 3 ------ (i)Subtract after multiplying the first by x and the second by 2.Thus (2x4 + x3 – 9x) - (2x4 + 4x2 + 18) = x3 - 4x2 – 9x – 18 ------ ( ii )Multiply (i) by x and subtract from (ii)x3 – 4x2 – 9x – 18 – (x3 + 2x2 + 3x) = - 6x2 – 12x – 18
14. 14. ÷ - 6 gives x2 + 2x + 3.Thus ( x2 + 2x + 3 ) is the H.C.F. of the given expressions.Find the H.C.F. in each of the following cases using Vedic sutras:1 x2 + 2x – 8, x2 – 6x + 82 x3 – 3x2 – 4x + 12, x3 – 7x2 + 16x - 123 x3 + 6x2 + 11x + 6, x3 – x2 - 10x - 84 6x4 – 11x3 + 16x2 – 22x + 8, 6x4 – 11x3 – 8x2 + 22x – 8.Edited byRakes Prasad , 2008Contains: (1) Use the formula ALL FROM 9 AND THE LASTFROM 10 to perform instant subtractions.(2) Multiplying numbers just over 100.(3) The easy way to add and subtract fractions.(4) Multiplying a number by 11.(5) Method for diving(6) SOME BASIC RELATIONS AND REVIEW OF SQUAREAND CUBE FORMULA:(7) SQUARE ROOTS AND MULTIPLICATION FORMULA :(8) TWO INTERESTING FACTS:(9) ROMAN NUMERALS(10) NUMER REPRESENTATION:(11) DECIMAL REPRENTATION AND PREFIXS:(12) POWER OF INTIGERS:(13) SOME PROOFS BUT WITHOUT WORDS:(14) THE LARGEST PRIME NUMBERS DISCOVERED SOFAR:Edited byRakes Prasad, 2008(1) Use the formula ALL FROM 9 AND THE LAST FROM 10 toperforminstant subtractions.● For example 1000 - 357 = 643 We simply take each figure in 357 from 9 and thelast figurefrom 10.So the answer is 1000 - 357 = 643And thats all there is to it!This always works for subtractions from numbers consisting of a 1 followed bynoughts: 100;1000; 10,000 etc.
15. 15. Similarly 10,000 - 1049 = 8951For 1000 - 83, in which we have more zeros than figures in the numbers beingsubtracted, wesimply suppose 83 is 083. So 1000 - 83 becomes 1000 - 083 = 917Try some yourself:1) 1000 - 777 2) 1000 - 283 3) 1000 - 505 4) 10,000 - 2345 5) 10000 - 9876 6)10,000 -1101(2) Multiplying numbers just over 100.● 103 x 104 = 10712The answer is in two parts: 107 and 12, 107 is just 103 + 4 (or 104 + 3), and 12 isjust 3 x 4.● Similarly 107 x 106 = 11342 107 + 6 = 113 and 7 x 6 = 42Again, just for mental arithmetic Try a few:1) 102 x 107 = 2) 106 x 103 = 1) 104 x 104 = 4) 109 x 108 =(3) The easy way to add and subtract fractions.Use VERTICALLY AND CROSSWISE to write the answer straight down!Multiply crosswise and add to get the top of the answer: 2 x 5 = 10 and 1 x 3 = 3.Then 10 + 3 =13. The bottom of the fraction is just 3 x 5 = 15..You multiply the bottom number together.Subtracting is just as easy: multiply crosswise as before, but the subtract:Try a few:(4) Multiplying a number by 11.To multiply any 2-figure number by 11 we just put the total ofthe two figures between the 2 figures.● 26 x 11 = 286Notice that the outerfigures in 286 are the26 being multiplied.And the middle figure is just 2 and 6 added up.● So 72 x 11 = 792Multiply by 11:1) 43 = ) 81 = 3) 15 = 4) 44 = 5) 11 =● 77 x 11 = 847This involves a carry figure because 7 + 7 = 14 we get 77 x 11 = 7147 = 847.Multiply by 11:1) 88 = 2) 84 = 3) 48 = 4) 73 = 5) 56 =● 234 x 11 = 2574
16. 16. We put the 2 and the 4 at the ends. We add the first pair 2 + 3 = 5. and we add thelast pair: 3 +4 = 7.Multiply by 11:1) 151 = 2) 527 = 3) 333 = 4) 714 = 5) 909 =(5) Method for diving by 9.23 / 9 = 2 remainder 5The first figure of 23 is 2, and this is the answer. The remainder is just 2 and 3added up!● 43 / 9 = 4 remainder 7The first figure 4 is the answer and 4 + 3 = 7 is the remainder - could it be easier?Divide by 9:1) 61 = remainder 2) 33 = remainder 3) 44 = remainder 4) 53 = remainder● 134 / 9 = 14 remainder 8The answer consists of 1,4 and 8.1 is just the first figure of 134.4 is the total of thefirst twofigures 1+ 3 = 4,and 8 is the total of all three figures 1+ 3 + 4 = 8.Divide by 9:6) 232 = 7) 151 = 8) 303 = 9) 212 = remainder remainder remainder remainder● 842 / 9 = 812 remainder 14 = 92 remainder 14Actually a remainder of 9 or more is not usually permitted because we are trying tofind howmany 9s there are in 842.Since the remainder, 14 has one more 9 with 5 left over the final answer will be 93remainder5Divide these by 9:1) 771 2) 942 3) 565 4) 555 5) 777 6) 2382(6) SOME BASIC RELATIONS AND REVIEW OF SQUARE AND CUBEFORMULA:Relations :If a = b , b = c then a= cIf a > b , b > c then a> cIf a < b , b < c then a < cIf a b< b c then a < cIf ab > bc then b > cIf a > b , c> d then a + c> b+ dIf a > b , c< d then a -c> b –dIf a < b , c< d then a -c< b -dIf a > b & a ,b both positive then 1/a < 1/bx = n then x = n or -n 2 2(a + b) = a + 2ab + b 2 2 2
17. 17. (a – b) = a –2ab + b 2 2 2(a + b) = a + 3a b + 3ab + b 3 3 2 2 3(a - b) = a -3a b + 3ab -b 3 3 2 2 3(a + b+ c) = a + b + c + 2ab + 2bc+ 2ac 2 2 2 2Factorizationa –b = ( a + b )(a –b) 2 2a + b = (a + b) ( a -ab + b ) 3 3 2 2a -b = (a -b ) (a + ab+ b ) 3 3 2 2Identities(a+ b) + (a –b) = 2(a + b ) 2 2 2 2(a+ b) -(a–b) = 4ab 2 2In dicesamxan=a(m+n)am/an=a(m-n)a=1 0a =1/a -mma = m√a 1/m(a *b ) = amx b m m(a /b ) = a mm/bm(a ) = a m n m*nLogarithmsa = n then log a n= x xlog a (mn )= log am + log anlog a (m/n) = log a m -log anlog a (m) = n log am nlog b n= log an/log abSurd( √a ) = a n nn√ a * √ b = √ab n nn√ a / √ b = √a /b n nm√ √ a = √a = √ √ a n mn n mm√ √ a = √a = √a p p mp p mAngle MeasurementTotal of Interior Angles in DegreesTriangle 180Rectangle 360Square 360Pentagon 540Circle 360Some FactsIn a triangle, interior opposite angle is always less than exterior angle.
18. 18. Sum of 2 interior opposite angles of a triangle is always equal to exterior angle.Triangle can have at one most obtuse angle.Angle made by altitude of a triangle with side on which it is drawn is equal to 90 degrees.in parallelogram opposite angles are equal.SquaresNumber Square Number Square Number Square1 1 11 121 21 4412 4 12 144 22 4843 9 13 169 23 5294 16 14 196 24 5765 25 15 225 25 6256 36 16 256 26 6767 49 17 289 27 7298 64 18 324 28 7849 81 19 361 29 84110 100 20 400 30 900Cubes & Other PowersNumber ( X) X3 X4 X511112 8 16 323 27 81 2434 64 256 10245 125 625 31256 216 1296 77767 343 2401 168078 512 4096 327689 729 6561 5904910 1000 10000 100000(7) SQUARE ROOTS AND MULTIPLICATION FORMULA :No Roots No Roots No Roots No Roots1 1 5 2.23 9 3 36 62 1.41 6 2.44 10 3.16 49 73 1.73 7 2.64 16 4 64 84 2 8 2.82 25 5 81 91 to 10 :1 2 3 4 5 6 7 8 9 101 1 2 3 4 5 6 7 8 9 102 2 4 6 8 10 12 14 16 18 203 3 6 9 12 15 18 21 24 27 304 4 8 12 16 20 24 28 32 36 405 5 10 15 20 25 30 35 40 45 506 6 12 18 24 30 36 42 48 54 607 7 14 21 28 35 42 49 56 63 708 8 16 24 32 40 48 56 64 72 809 9 18 27 36 45 54 63 72 81 9010 10 20 30 40 50 60 70 80 90 10011to 20
19. 19. 11 12 13 14 15 16 17 18 19 201 11 12 13 14 15 16 17 18 19 202 22 24 26 28 30 32 34 36 38 403 33 36 39 42 45 48 51 54 57 604 44 48 52 56 60 64 68 72 76 805 55 60 65 70 75 80 85 90 95 1006 66 72 78 84 90 96 102 108 114 1207 77 84 91 98 105 112 119 126 133 1408 88 96 104 112 120 128 136 144 152 1609 99 108 117 126 135 144 153 162 171 18010 110 120 130 140 150 160 170 180 190 200(8) TWO INTERESTING FACTS:(9) ROMAN NUMERALS(10) NUMER REPRESENTATION:(11) DECIMAL REPRENTATION AND PREFIXS:(12) POWER OF INTIGERS:(13) SOME PROOFS BUT WITHOUT WORDS:(14) THE LARGEST PRIME NUMBERS DISCOVERED SO FAR:THE ENDEdited byRakes Prasad ,2008Contains: 1. Some different techniques of squaringending with 1,2,3,4,6,7,8,9,etc .2. adding different number sequence3. finding the difference of squares.4. dividing a 6-digit/3-digit number by13,7,37037,11,41,15873 etc5. dividing numbers by 75,125,625,12 2/3 etc..6. know why divisibility rules works?7. finding percentage quickly.8. various types of multiplication by99,72,84,………9. squaring of some specialnumbersRakes Prasad2008.Squaring a 2-digit number ending in 1
20. 20. .Take a 2-digit number ending in 1. Subtract 1 from thenumber. Squarethe difference. Add the difference twice to its square. Add 1.Example:If the number is 41, subtract 1: 41 - 1 = 40. 40 x 40 = 1600(square thedifference). 1600 + 40 + 40 = 1680 (add the difference twice toits square).1680 + 1 = 1681 (add 1). So 41 x 41 = 1681.See the pattern?For 71 x 71, subtract 1: 71 - 1 = 70.70 x 70 = 4900 (square the difference).4900 + 70 + 70 = 5040 (add the difference twice to itssquare). . So 71 x 71= 5041.Squaring a 2-digit number ending in 2I Take a 2-digit number ending in 2. f The last digit will be _ __ 4. tMultiply the first digit by 4: the 2nd number will be h the nextto the lastdigit: _ _ X 4. e n Square the first digit and add the numbercarried from uthe previous step: X X _ _. m b e Example: r is 52, the last digit is _ _ _ 4. 4 x 5 = 20 (four times the firstdigit): _ _ 0 4. 5x 5 = 25 (square the first digit), 25 + 2 = 27 (add carry): 2 7 0 4.For 82 x82, the last digit is _ _ _ 4. 4 x 8 = 32 (four times the firstdigit): _ _ 2 4. 8 x8 = 64 (square the first digit), 64 + 3 = 67 (add carry): 6 7 2 4.Squaring a 2-digit number ending in 3 Take a 2-digitnumberending in 3. 2 The last digit will be _ _ _ 9. 3 Multiply the firstdigit by 6:
21. 21. the 2nd number will be the next to the last digit: _ _ X 9. 4Square the firstdigit and add the number carried from the previous step: X X_ _.Example:1 If the number is 43, the last digit is _ _ _ 9. 2 6 x 4 = 24 (sixtimes thefirst digit): _ _ 4 9. 3 4 x 4 = 16 (square the first digit), 16 + 2 =18 (addcarry): 1 8 4 9.4 So 43 x 43 = 1849.See the pattern?1 For 83 x 83, the last digit is _ _ _ 9. 2 6 x 8 = 48 (six timesthe first digit):_ _ 8 9.38 x 8 = 6 4 ( s q u a r e t he first digit), 64 + 4 = 68 (add carry):6 8 8 9. 4So 83 x 83 = 6889.Squaring a 2-digit number ending in 41 Take a 2-digit number ending in 4. 2 Square the 4; the lastdigit is 6: _ __ 6 (keep carry, 1.)3 Multiply the first digit by 8 and add the carry (1); the 2ndnumber will bethe next to the last digit: _ _ X 6 (keep carry). 4 Square thefirst digit andadd the carry: X X _ _.Example:1 If the number is 34, 4 x 4 = 16 (keep carry, 1); the last digitis _ _ _ 6. 2 8x 3 = 24 (multiply the first digit by 8), 24 + 1 = 25 (add thecarry): the nextdigit is 5: _ _ 5 6. (Keep carry, 2.) 3 Square the first digit andadd the
26. 26. 1 If the two numbers selected are 6 and 19: 2 Add thenumbers: 6 + 19 =25. 3 Subtract the numbers: 19 - 6 = 13. Add 1: 13 + 1 = 14. 4Multiply 25by half of 14: 25 x 7 = 175. 5 So the sum of the numbers from6 through 19is 175.Therefore 6+7+8+9+10+11+12+13+14+15+16+17+18+19=175See the pattern?1 If the two numbers selected are 4 and 18: 2 Add thenumbers: 4 + 18 =22. 3 Subtract the numbers: 18 - 4 = 14. Add 1: 14 + 1 = 15.Multiply half of 22 by 15: 11 x 15 = 165 (10 x 15 + 15). So thesum of thenumbers from 4 through 18 is 165.1 Choose a 2-digit odd number. Add all the odd numbersstarting withone through this 2-digit number: 2 Add one to the 2-digitnumber. 3Divide this sum by 2 (take half of it). 4 Square this number.This is thesum of all odd numbers from 1 through the 2-digit numberchosen.Example:1 If the 2-digit odd number selected is 35: 2 35+1 = 36 (add 1).3 36/2 = 18(divide by 2) or 1/2 x 36 = 18 (multiply by 1/2). 4 18 x 18 = 324(square 18):18 x 18 = (20 - 2)(18) = (20 x 18) - (2 x 18) = 360 - 36 = 360 -30 -6 = 324. 5So the sum of all the odd numbers from one through 35 is324.See the pattern?1 If the 2-digit odd number selected is 79: 2 79+1 = 80 (add 1).3 80/2 = 40
27. 27. (divide by 2) or 1/2 x 80 = 40 (multiply by 1/2). 4 40 x 40 = 1600(square40).5. So the sum of all the odd numbers from one through 79 is1600.1 Have a friend choose a a single digit number. (Norestrictions forexperts.) 2 Ask your friend to jot down a series of doubles(where thenext term is always double the preceding one), and tell youthe last term.3 Ask your friend to add up all these terms. 4 You will givethe answerbefore he or she can finish: The sum of all the terms of thisseries will betwo times the last term minus the first term.Example:if the number selected is 9: 1 The series jotted down is: 9, 18,36, 72, 144.2 Two times the last term (144) minus the first (9): 2 x 144 =288; 288 - 9 =279.3 So the sum of the doubles from 9 through 144 is 279.See the pattern? Heres one for the experts:1 The number selected is 32: 2 The series jotted down is: 64,128, 256,512. 3 Two times the last term (512) minus the first (64): 2 x512 = 1024;1024 - 32 = 1024 - 30 - 2 = 994 - 2 = 992.4 So the sum of the doubles from 32 through 512 is 992.Remember to subtract in steps from left to right. Withpractice you will beexpert in summing series.restrictions for experts.) Ask your friend to jot down a seriesof
28. 28. quadruples (where the next term is always four times thepreceding one),and tell you only the last term. Ask your friend to add up allthese terms.You will give the answer before he or she can finish: Thesum of all theterms of this series will be four times the last term minus thefirst term,divided by 3. Example:If the number selected is 5:1 The series jotted down is: 5, 20, 80, 320, 1280. 2 Four timesthe last term(1280) minus the first (5): 4000 + 800 + 320 - 5 = 5120 - 5 =5115 Divide by3: 5115/3 = 17053 So the susum of the quadruples from 5 through 1280 is1705.See the pattern? Heres one for the experts:1 The number selected is 32: 2 The series jotted down is: 32,128, 512,2048. 3 Four times the last term (2048) minus the first (32):8000 + 160 +32 - 32 = 8,160Divide by 3: 8160/3 = 2720.4. So the sum of the quadruples from 32 through 2048 is2720.Practice multiplying from left to right and dividing by 3. Withpractice youwill be anexpert quad adder.Add a sequence from one to a selected 2-digit number1 Choose a 2-digit number. 2 Multiply the 2-digit number byhalf the nextnumber, orMultiply half the 2-digit number by the nextnumber.
29. 29. Example:1 If the 2-digit even number selected is 51: 2 The nextnumber is 52.Multiply 51 times half of 52. 3 51 x 26: (50 x 20) + (50 x 6) + 1 x26) = 1000+ 300 + 26 = 13264 So the sum of all numbers from 1 through 51 is 1326.See the pattern?1 If the 2-digit even number selected is 34: 2 The nextnumber is 35.Multiply half of 34 x 35. 3 17 x 35: (10 x 35) + (7 x 30) + (7 x 5)= 350 + 210+ 35 = 560 + 35 = 5954 So the sum of all numbers from 1 through 34 is 595.With some multiplication practice you will be able to findthese sums ofsequential numbers easily and faster than someone using acalculator!1 Choose a 1-digit number. 2 Square it.Example:1 If the 1-digit number selected is 7: 2 To add 1 + 2 + 3 + 4 + 5+6+7+6+5 + 4 + 3 + 2 + 1 3 Square 7: 49 4 So the sum of all numbersfrom 1through 7 and back is 49.See the pattern?1 If the 1-digit number selected is 9: 2 To add 1 + 2 + 3 + 4 + 5+6+7+8+9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 3 Square 9: 81 4 So the sum ofall numbersfrom 1 through 9 and back is 81.Add sequences of numbers in the 10s1 Choose a 2-digit number in the 10s. To add all the 10sfrom 10 up
30. 30. through this number and down from it: 2 Square the 2nd digitof thenumber (keep the carry) _ _ X 3 The number of terms is 2 xthe 2nd digit +1. 4 The first digits = the number of terms (+ carry). nbsp; X X_Example:1 If the 2-digit number in the 10s selected is 16: (10 + 11 +12 ... 16 + 15 +14 ... 10) 2 Square the 2nd digit of the number: 6 x 6 = 36(keep carry 3) _ _6 3 No. of terms = 2 x 2nd digit + 1: 2 x 6 + 1 = 134 No. of terms (+ carry): 13 + 3 = 16 1 6 _ 5 So the sum of thesequence is166.See the pattern?1. If the 2-digit number in the 10s selected is 18:(10 + 11 + 12 + ... 18 + 17 + 16 + 15 ... 10)1 Square the 2nd digit of the number: 8 x 8 = 64 (keep carry6) _ _ 4 2 No.of terms = 2 x 2nd digit + 1: 2 x 8 + 1 = 17 3 No. of terms (+carry): 17 + 6 =23 2 3 _ 4 So the sum of the sequence is 234.1 Choose a 2-digit number in the 20s. To add all the 20sfrom 20 upthrough this number and down from it: 2 Square the 2nd digitof thenumber (keep the carry) _ _ X 3 The number of terms is 2 xthe 2nd digit +1. 4 Multiply the number of terms by 2 (+ carry). X X _Example:1 If the 2-digit number in the 20s selected is 23: (20 + 21 + 22+ 23 + 22 +21 + 20) 2 Square the 2nd digit of the number: 3 x 3 = 9 _ _ 9 3No. of
31. 31. terms = 2 x 2nd digit + 1: 2 x 3 + 1 = 7 4 2 x no. of terms: 2 x 7= 14 1 4 _ 5So the sum of the sequence is 149.See the pattern?1 If the 2-digit number in the 20s selected is 28: (20 + 21 +22 ... + 28 + 27+ ... 22 + 21 + 20) 2 Square the 2nd digit of the number: 8 x 8= 64 (keepcarry 6) _ _ 4 3 No. of terms = 2 x 2nd digit + 1: 2 x 8 + 1 = 174 2 x no. ofterms (+ carry): 2 x 17 + 6 = 40 4 0 _ 5 So the sum of thesequence is 404.2 S 1 Choose a 2-digit number in the 30s. To add all the 30sqfrom 30 upthrough this number and down from it: uare the 2nd digit of the number (keep the carry) _ _ X 3 Thenumber ofterms is 2 x the 2nd digit + 1. 4 Multiply the number of termsby 3 (+carry) X X _Example:1 If the 2-digit number in the 30s selected is 34: (30 + 31 + 32+ 33 + 34 +33 + 32 + 31 + 30) 2 Square the 2nd digit of the number: 4 x 4= 16 (keepcarry 1) _ _ 6 3 No. of terms = 2 x 2nd digit + 1: 2 x 4 + 1 = 94 3 x no. of terms: 3 x 9 + 1 = 28 2 8 _ 5 So the sum of thesequence is286.See the pattern?1 If the 2-digit number in the 30s selected is 38: (30 + 31 + 232 + ... + 38 +37 + ... 32 + 31 + 30) 3 Square the 2nd digit of the number: 8 x8 = 64 (keepcarry 6) _ _ 4
32. 32. No. of terms = 2 x 2nd digit + 1: 2 x 8 + 1 = 173 x no. of terms: 3 x 17 + 6 = 51 + 6 = 57 5 7 _ So the sum ofthe sequenceis 574.1 Choose a 2-digit number in the 40s. To add all the 40sfrom 40 up through this number and down from it:2 Square the 2nd digit of the number(keep the carry) _ _X3 The number of terms is 2 x the 2nd digit + 1.4 Multiply the number of terms by 4 (+ carry) X X _e 2-digit number in the 40s selected is 43: (40 + 41 + 42 + 43+ 42 + 41 +40) 1 Square the 2nd digit of the number: 3 x 3 = 9 _ _ 9 2 No.of terms = 2x 2nd digit + 1: 2 x 3 + 1 = 73 4 x no. of terms: 4 x 7 = 28 2 8 _ 4 So the sum of thesequence is 289.See the pattern?1 If the 2-digit number in the 40s selected is 48: (40 + 41 + 42+ ... + 48 +47 + ... 42 + 41 + 40) 2 Square the 2nd digit of the number: 8 x8 = 64 (keepcarry 6) _ _ 4 3 No. of terms = 2 x 2nd digit + 1: 2 x 8 + 1 = 174 4 x no. ofterms: 4 x 17 + 6 = 40 + 28 + 6 = 68 + 6 = 74 7 4 _5 So the sum of the sequence is 744.1 Choose a 2-digit number in the 50s. To add all the 50sfrom 50 upthrough this number and down from it: 2 Square the 2nd digitof thenumber (keep the carry) _ _ X 3 The number of terms is 2 xthe 2nd digit +1. 4 Multiply the number of terms by 5. X X _Example:
33. 33. 1 If the 2-digit number in the 50s selected is 53: (50 + 51 + 52+ 53 + 52 +51 + 50) 2 Square the 2nd digit of the number: 3 x 3 = 9 _ _ 9 3No. ofterms = 2 x 2nd digit + 1: 2 x 3 + 1 = 7 4 5 x no. of terms: 5 x 7= 35 3 5 _ 5So the sum of the sequence is 359.See the pattern?1 If the 2-digit number in the 50s selected is 57: (50 + 51 + 52 ... + 57 + 56 + ... 52 + 51 + 50) 2 Square the 2nd digit of thenumber: 7 x 7= 49 (keep carry 4) _ _ 93 No. of terms = 2 x 2nd digit + 1: 2 x 7 + 1 = 155 x no. of terms (+ carry): 5 x 15 + 4 = 75 + 4 = 79 7 9 _So the sum of the sequence is 799.Add sequences of numbers in the 60s1 Choose a 2-digit number in the 60s. To add all the 60sfrom 60 up throughthis number and down from it: 2 Square the 2nd digit of thenumber (keepthe carry) _ _ X 3 The number of terms is 2 x the 2nd digit + 1.4 Multiply thenumber of terms by 6 (+ carry) X X _Example:1 If the 2-digit number in the 60s selected is 63: (60 + 61 + 62+ 63 + 62 + 61 +60) 2 Square the 2nd digit of the number: 3 x 3 = 9 _ _ 9 3 No.of terms = 2 x2nd digit + 1: 2 x 3 + 1 = 745 6 x no. of terms: 6 x 7 = 42 4 2 _ So the sum of thesequence is 429.See the pattern?1 If the 2-digit number in the 60s selected is 68: (60 + 61 +62 + ... + 68 + 67 + ... 62 + 61 + 60) 2 Square the 2nd digit of
34. 34. the number: 8 x 8 = 64 (keep carry 6) _ _ 4 3 No. of terms =2 x 2nd digit + 1: 2 x 8 + 1 = 17 4 6 x no. of terms: 6 x 17 +6 = 42 + 1 = 102 + 6 = 108 1 0 8 _5oSthe sum of the sequence is 1084.1 Choose a 2-digit number in the 70s. To add all the 70sfrom 70 upthrough this number and down from it: 2 Square the 2nd digitof thenumber (keep the carry) _ _ X3 The number of terms is 2 x the 2nd digit + 1. 4 Multiply thenumber ofterms by 7 (+ carry) X X _Example:1 If the 2-digit number in the 70s selected is 73: (70 + 71 + 72+ 73 + 72 +71 + 70) 2 Square the 2nd digit of the number: 3 x 3 = 9 _ _ 9 3No. ofterms = 2 x 2nd digit + 1: 2 x 3 + 1 = 74 7 x no. of terms: 7 x 7 = 49 4 9 _ 5 So the sum of thesequence is 499.See the pattern?1 If the 2-digit number in the 70s selected is 78: (70 + 71 + 72+ ... + 78 +77 + ... 72 + 71 + 70) 2 Square the 2nd digit of the number: 8 x8 = 64 (keepcarry 6) _ _ 4 3 No. of terms = 2 x 2nd digit + 1: 2 x 8 + 1 = 174 7 x no. ofterms: 7 x 17 + 6 = 49 + 6 = 119 + 6 = 125 1 2 5 _5 So the sum of the sequence is 1254.1 Choose a 2-digit number in the 80s. To add all the80sfrom 80 up through this number and down from it:2 Square the 2nd digit of the number(keep the carry) _ _
35. 35. umber of terms is 2 x the 2nd digit + 1 4 Multiply the numberof terms by8 (add the carry) X X _Example:1 If the 2-digit number in the 80s selected is 83: (80 + 81 + 82+ 83 + 82 +81 + 80) 2 Square the 2nd digit of the number: 3 x 3 = 9 _ _ 9 3No. ofterms = 2 x 2nd digit + 1: 2 x 3 + 1 = 7 4 8 x no. of terms: 8 x 7= 56 5 6 _ 5So the sum of the sequence is 569.See the pattern?1 If the 2-digit number in the 80s selected is 88: (80 + 81 +82 ... + 88 + 87+ ... 83 + 82 + 81 + 80) 2 Square the 2nd digit of the number: 8x 8 = 64(keep carry 6) _ _ 4 3 No. of terms = 2 x 2nd digit + 1: 2 x 8 + 1= 178 x no. of terms (+ carry): 8 x 17 + 6 = 80 + 56 + 6 = 136 + 6 =42 4 2 _So the sum of the sequence is 1424.1 Have a friend choose and write down a single-digit number.(Twodigits for experts.) 2 Ask your friend to name and note a thirdnumberby adding the first two. 3 Name a fourth by adding thesecond and third.Continue in this way, announcing each number, through tennumbers. 4Ask your friend to add up the ten numbers. You will give theanswerbefore he or she can finish:The sum of all the terms of this series will be the seventhnumbermultiplied by 11.
36. 36. Example:1 If the numbers selected are 7 and 4:e series jotted down is: 4, 7, 11, 18, 29, 47, 76, 123, 199, 322. 3Theseventh number is 76. 11 x 76 = 836 (use the shortcut for 11:7 is the firstdigit, 6 is the third digit; the middle digit will be 7 + 6, andcarry the 1:836).4 So the sum of the ten numbers is 836.Here are some of the calculations that you can do mentallyafterpracticing the exercises. See how many you can do. Checkyour answerswith a calculator. Your mental math powers should beimpressive!15 x 15 89 x 89 394 x 101 32 x 38 51 x 59101 x101 x 94 93 x 93 228 x 101 25 x 25147 56 x 56 94 x 96 101 x 448 83 x 87 687 x 101 64 x 66 52 x 52101 x 65435 x 35 11 x 19 101 x61 x 69 89 x 101 61 x 61 34 x 36206 101 x 48 52 x 52 45 x 45 456 x 101 54 x 54 41 x 41 101 x 4132 x 32 83 x101 29 x 29 55 x 55 63 x 67 479 x 101 82 x 82 882 x 101 14 x 16319 x 10173 x 77 65 x 65 101 x 859 13 x 17 101 x 149 41 x 49 82 x 101 75x 75 101 x71 x 79 69 x 101 22 x 22 993 x 101144 51 x 51 85 x 85 101 x 101 129 x 101 69 x 69 31 x 39 738 x101 21 x 2971 x 71 77 x 101 95 x 95 265 x 101 74 x 76 53 x 53 101 x 409 51x 51 98 x101 94 x 94 42 x 42 53 x 57 339 x
37. 37. 62 x 68 101 x 88 19 x 19 238 x 10110143 x 47 96 x 96 21 x 21 101 x 279 81 x 89648 x62 x 62 101 x 57 24 x 26 101 x 668101 99 x 99 23 x 27 22 x 28 515 x 101 97 x 97 79 x 79 101 x 826245 x 10131 x 31 37 x 33 101 x82 x 88 39 x 39 126 x 101 91 x 9921868 x 101 51 x 101 598 x 101 98 x 98 72 x 7291 x 91 54 x 56 101 x 29 57 x 57 771 x 101 81 x 81 559 x 101 52x 58 59 x 5984 x 86 46 x 44 101 x 25 12 x 18 101 x 697 92 x 92 349 x93 x 97 101 x 188 42 x 48 49 x 4910158 x 58 92 x 92 101 x 78 37 x 1011 Select two consecutive 2-digit numbers. 2 Add the two 2-digit numbers!Examples:1 24 + 25 = 49. (Try it on a calculator and see, or if yourereally sharp, doit mentally: 24 x 24 = 576, 25 x 25 = 625, 625 - 576 = 49.) 2 If 63and 64 areselected, then 63 + 64 = 127. (For larger number addition, doit in steps:63 + 64 = 63 + 60 + 4 = 123 + 4 = 127.)1 Select two consecutive 2-digit numbers, one not more than10 largerthan the other (experts need not use this limitation). 2Subtract thesmaller number from the larger. 3 Add the two numbers. 4Multiply thefirst answer by the second.Examples:
38. 38. 1 If 71 and 64 are selected: 2 71 - 64 = 7. 3 71 + 64 =Add left to right: 71 + 64 = 71 + 60 + 4 = 131 + 4 = 135)4 Multiply these results: 7 x 135 = 945 (Multiply left to right: 7x 135 = 7 x(100+30+5) = 700 + 210 + 35 = 910 + 35 = 945)5 So the difference of the squares of 71 and 64 is 945.See the pattern?1 If 27 and 36 are selected: 3 2 36 - 27 = 9.36 + 27 = 63 (Think: 27 + 30 + 6 = 57 + 6 = 63)Multiply these results: 9 x 63 = 567 (Think: 9 x (60+3) = 540 +27 = 567)So the difference of the squares of 27 and 36 is 567.Why do the divisibility rules work?From: cms@dragon.com (Cindy Smith) To: Dr. Math<dr.math@forum.swarthmore.edu> Subject: Factoring TricksIn reading a popular math book, I came across severalarithmeticfactoring tricks. Essentially, if the last digit of a number iszero,then the entire number is divisible by 10. If the last number iseven, then the entire number is divisible by 2. If the last twodigits are divisible by 4, then the whole number is. If the lastthree digits divide by 8, then the whole number does. If thelastfour digits divide by 16, then the whole number does, etc. Ifthelast digit is 5, then the whole number divides by 5.Now for the tricky ones.If you add the digits in a number and the sum is divisible by3,then the whole number is. Similarly, fif you add the digits ina number and the sum is divisible by 9, then the wholenumber is. For example, take the number 1233: 1 + 2 + 3 + 3 =9.
39. 39. Therefore, the whole number is divisible by 9 and thequotient is137. The number is also divisible by 3 and the quotient is411.It works for extremely large numbers too (I checked on mycalculator). Now heres a really tricky trick. You add up thealternate digits of a number and then add up the other setof alternate digits. If the sums of the alternate digits equaleach other then the whole number is divisible by 11. Also,if the difference of the alternate digits is 11 or a multiple of11, then the whole number is divisible by 11. For example,123,456,322. 1 + 3 + 5 + 3 + 2 = 14 and 2 + 4 + 6 + 2 = 14.Therefore, the whole number is divisible by 11 and thequotientis 11,223,302. Also, if a number is divisible by both 3 and 2,thenthe whole number is divisible by 6.The only single digit number for which there is no trick listedis 7.I find these rules interesting and useful, especially whenfactoringlarge numbers in algebraic expressions. However, Im notsurewhy all these rules work. Can you explain to me why thesemathtricks work? If I could understand why they work, I think itwouldimprove my math skills. Thanks in advance for your help.Cindy Smith cms@dragon.comFrom: Dr. Math <dr.math@forum.swarthmore.edu> To:cms@dragon.com (Cindy Smith) Subject: Re: Factoring TricksThank you for this long and very well-written question. I willtry towrite as clearly as you as I answer. All these digital tests for
40. 40. divisibility are based on the fact that our system of numeralsiswritten using the base of 10.The digits in a string of digits making up a numeral areactually thecoefficients of a polynomial with 10 substituted for thevariable. Forexample, 1233 = 1*10^3 + 2*10^2 + 3*10^1 + 3*10^0 which isgottenfrom the polynomial 1*x^3 + 2*x^2 + 3*x^1 + 3*x^0 = x^3 +2*x^2 +3*x + 3 by substituting 10 for x. We can explain each of thesetricksin terms of that fact: 10 - numerals ending in 0 representnumbersdivisible by 10:Since the last digit is zero, and all other terms in thepolynomial formare divisible by 10, the number is divisible by 10. Similarly, ifthenumber is divisible by 10, since all the terms except the lastoneare automatically divisible by 10 no matter what thecoefficients ordigits are, the number will be divisible by 10 only if the lastdigit is.Since all the digits are smaller than 10, the last digit has to be0 tobe a multiple of 10.2 - numerals ending in an even digitrepresentnumbers divisible by 2: f o Same argument as above about alltheterms except the last one being divisible by 2. The last digitis
41. 41. divisible by 2 (even) if and only if the whole number is. R 4 - numerals endingwitha two-digit multiple of 4 represent numbers divisible by 4:o fSimilar to the above, but since 10 is not a multiple of 4, but10^2 is,we have to look at the last two digits instead of just the lastdigit.a n 8 - numerals ending with a three-digit multiple of 8representnumbers divisible by 8: u Similar to 4, but now 10^2 is not amultipleof 8, but 10^3 is, so we have to look at the last three mdigits.era16 - You figure this one!l5 - You figure this one, too!ook3 - numerals whose sum of digits is divisible by 3 representnumberss divisible by 3: This one is different, because 3 does notdivide anypower of 10 evenly. That means we will have to consider theeffect ofall the l digits. Here we use this fact: 10^k - 1 = (10 - 1)*(10^(k-1) = ... +10^2 + i 10 + 1) This is a fancy way of saying 9999...999 =9*1111...111. We use k this to rewrite our powers of 10 as 10^k = 9*a[k] +1. Now thepolynomial e this: d[k]*10^k + d[k-1]*10^(k-1) + ... + d[1]*10 +d[0] = d[k]*(9*a[k] + 1) + ... + d[1]*(9*a[1] + 1) + d[0] = 9*(d[k]*a[k] + ... +d[1]*a[1])
42. 42. + d[k] + ... + d[1] + d[0] Now notice that 3 divides the firstpart, so thewhole number is divisible by 3 if and only if the sum of the digits is.9 - numerals whose sum of digits is divisible by 9 representnumbersdivisible by 9: Use the same equation as the previous case.You figurethe rest!11 - numerals whose alternating sum of digits is divisible by11 representnumbers divisible by 11:Here the phrase "alternating sum" means we alternate thesigns frompositive to negative to positive to negative, and so on. Weuse this fact:10 to an odd power plus 1 is divisible by 11, and 10 to aneven powerminus 1 is divisible by 11. The first part is a fancy way ofwriting 10000...0001 = 11*9090...9091 (where there are an even number of0s on theleft-hand side). The second part is a fancy way of writing99999...9999= 11*9090...0909 (where there are an even number of 9s inthe lefthand side). We write 10^(2*k) = 11*b[2*k] + 1 and 10^(2*k+1) =11*b[2*k+1] - 1. Here k is any nonnegative integer. We substitutethatinto the polynomial form, so:d[2*k]*10^(2*k) + d[2*k-1]*10^(2*k-1) + ... + d[1]*10 + d[0]= d[2*k]*(11*b[2*k] + 1) + d[2*k-1]*(11*b[2*k-1] - 1) + ...
43. 43. � + d[1]*(11*a[1] - 1) + d[0] = 11*(d[2*k]*b[2*k] + ... + d[1]*a[1])+ d[2*k]- d[2*k-1]� + ... - d[1] + d[0]The first part is divisible by 11 no matter what the digits are,so thewhole number is divisible by 11 if and only if the last part,which isthe alternating sum of the digits, isdivisible by 11. If you prefer, you can writed[2*k] - d[2*k-1] + ... - d[1] + d[0] = (d[0] + d[2] + ... + d[2*k]) -(d[1] +d[3] + ... + d[2*k-1]), so that you add up every other digit,starting fromthe units digit, and then add up the remaining digits, andsubtract thetwo sums. This will compute the same result as thealternating sum ofthe digits. 7 - There is a trick for 7 which is not as well knownas theothers. It makes use of the fact that 10^(6*k) - 1 is divisible by7, and10^(6*k - 3) + 1 is divisible by 7. It goes like this: Mark off thedigits ingroups of threes, just as you do when you put commas inlarge numbers.Starting from the right, compute the alternating sum of thegroups asthree-digit numbers. If the result is negative, ignore the sign.If theresult is greater than 1000, do the same thing to the resultingnumberuntil you have a result between 0 and 1000 inclusive. That 3-digit
44. 44. number is divisible by 7 if and only if the original number istoo.Example:123471023473 = 123,471,023,473, so make the sum473 - 23 + 471 - 123 = 450 + 348 = 798.798 = 7*114, so 798 is divisible by 7, and 123471023472 is,too.An extra trick is to replace every digit of 7 by a 0, every 8 by a1, andevery 9 by a 2, before, during, or after the sum, and the factremains.The sum could also have been computed as473 - 23 + 471 - 123 --> 403 - 23 + 401 - 123 = 380 + 278 --> 310+ 201 =511 = 7*73. You can figure out why this "casting out 7s" partworks.There is another way of testing for 7 which uses the fact that7 divides2*10 + 1 = 21. Start with the numeral for the number you wantto test.Chop off the last digit, double it, and subtract that from therest of thenumber. Continue this until you get stuck. The result is 7, 0,or -7, ifand only if the original number is a multiple of 7.Example:123471023473 --> 12347102347 - 2*3 = 12347102341 -->1234710234- 2*1 = 1234710232 --> 123471023 - 2*2 = 123471019 -->12347101 - 2*9= 12347083 --> 1234708 - 2*3 = 1234702--> 123470 - 2*2 = 123466 --> 12346 - 2*6 = 12334 --> 1233 - 2*4= 1225--> 122 - 2*5 = 112 --> 11 - 2*2 = 7.
45. 45. 13 - The same trick that works for 7 works for 13; that is, 13divides10^(6*k) - 1 and 10^(6*k - 3) + 1, so the alternating sum ofthree-digitgroups works here, too. 17 - This is harder. You would havetouse alternating sums of 8-digit groups!1 Select a 3-digit number. 2 Repeat these digits to make a 6-digitnumber. 3 Divide these 6 digits by 7, then by 13. 4 Theanswer is 11 times the first three digits!Example:1 If the 3-digit number selected is 234: 2 The 6-digit numberis 234234.3. Divide by 7, then by 13: multiply by 11-to multiply 234 by 11, work right to left: last digit on right = ___4next digit to left = 3 + 4 = 7: _ _ 7 _ next digit to left = 2 + 3 =5: _ 5 __ last digit on left = 2 _ _ _3 So 234234 divided by 7, then 13 is 2574.See the pattern?1 If the 3-digit number selected is 461: 2 The 6-digit numberis 461461.3. Divide by 7, then by 13: multiply by 11-to multiply 461 by 11, work right to left: last digit on right = ___1nextdigit to left = 4 + 6 = 10: _ 0 _ _ last digit on left = 4 + 1(carry) = 5: 5 _ _ _3 So 461461 divided by 7, then 13 is 5071.Practice multiplying by 11 - this process works formultiplying anynumber by 11.
46. 46. 1 Select a 3-digit number. 2 Repeat these digits to make a 6-digitnumber. 3 Divide these 6 digits by 13, then by 11. 4 Theansweris 7 times the first three digits!Example:1 If the 3-digit number selected is 231: 2 The 6-digit numberis 231231.3 Divide by 13, then by 11:7 x 231 = 1400 + 210 + 7 = 1617.4 So 231231 divided by 13, then 11 is 1617.See the pattern?1 If the 3-digit number selected is 412: 2 The 6-digit numberis 412412.Remembertomultiply left to right and add in increments. Then you willbe able to give these answers quickly and accurately.1 Select a 6-digit number repeating number. 2 Repeat thesedigits tomake a 6-digit number. 3 Multiply a single digit by 3, then by5.Example:1 If the 6-digit repeating number selected is 333333: 2Multiply 3 x 3: 9 3Multiply 9 x 5: 45 4 So 333333 divided by 37037 andmultiplied by 5 is 45.See the pattern?You can expand this exercise by using a different number inthe final step.Example: multiply by 4:1 If the 6-digit repeating number selected is 555555: 2Multiply 3 x 5: 15. 3
47. 47. Multiply 15 x 4: 60 4 So 555555 divided by 37037 andmultiplied by 4 is 60.By changing the last step you can generate many extensionsof this3 dsivide by 13, then by 11:7 x 412 = 2800 + 70 + 14 = 2884.4 So 412412 divided by 13, then 11 is 2884.exercise. Be inventive and create some impressivecalculations.Dividing a repeating 6-digit number by 7, 11, 13; subtract 101Select a 3-digit number.Repeat these digits to make a 6-digit number.Divide this 6-digit by 7, then 11, then 13.Subtract 101.The answer is the original number minus 101!Example:If the 3-digit number selected is 289:The 6-digit number is 289289.Divide by 7, then by 11, then by 13:the answer is 289.Subtract 101: 289 - 101 = 188So 289289 divided by 7, then 11, then 13 minus 101 is 188.See the pattern?If the 3-digit number selected is 983:The 6-digit number is 983983.Divide by 7, then by 11, then by 13:the answer is 983.1 Subtract 101: 983 - 101 = 882So 983983 divided by 7, then 11, then 13 minus 101 is 882.1 Select a repeating 3-digit number . 2 The answer is 3 timesoneof the digits plus 41!Example:1 Select 999. 2 Multiply one digit by 3: 9 x 3 = 27. 3 Add 41: 27+ 41
48. 48. = 68. 4 So (999 / 37) + 41 = 68.Change the last step to add other numbers, and thusproduce manynew exercises.1 Select a repeating 6-digit number . 2 The answer is 7 timesthe firstdigit of the number!Example:1 777777 / 15873 = 7 x 7 = 49. 2 555555 / 15873 = 7 x 5 = 35. 3999999 /15873 = 7 x 9 = 63. Not very demanding mental math, butgood for aquick challenge or two.Dividing mixed numbers by 21 Select a mixed number (a whole number and a fraction).� 2. If the whole number is even, divide by 2 - this is thewhole numberof the answer.�� The numerator of the fraction stays the same; multiplythedenominator by 2.� 3. If the whole number is odd, subtract 1 and divide by 2 -thisis the whole number of the answer.� Add the numerator and the denominator of the fraction -thiswill be the new numerator of the fraction;� Multiply the denominator by 2.Even whole number example:If the first number selected is 8 3/4: Divide the whole number(8) by
49. 49. 2: 8/2 = 4 (whole number) Use the same numerator: 3 Multiplythedenominator by 2: 4 x 2 = 8 (denominator) So 8 3/4 divided by2 = 4 3/8.Odd whole number example:1 If the first number selected is 13 2/5: 2 3 Subtract 1 from thewholenumber and divide by 2: 13 - 1 = 12, 12/2 = 6.Add the numerator and the denominator: 2 + 5 = 7. This is thenumerator of the fraction. Multiply the denominator by 2: 5 x2 = 10. So 13 2/5 divided by 2 = 6 7/10.file:///1 Select a 2-digit number. 2 Multiply it by 8 (or by 2 threetimes).3 Move the decimal point 2 places to the left.Example:1 The 2-digit number chosen to multiply by 12 1/2 is 78. 2Multiplyby 2 three times: 2 x 78 = 156 2 x 156 = 312 2 x 312 = 6243 Move the decimal point 2 places to the left: 6.24 4 So 78dividedby 12 1/2 = 6.24.See the pattern?1 If the 2-digit number chosen to multiply by 12 1/2 is 91: 2Double three times: 182, 364, 728. 3 Move the decimal point2 places to the left: 7.28 4 So 91 divided by 12 1/2 = 7.28.5ividing a 2-digit number by 151 Select a 2-digit number. 2 Multiply it by 2. 3 Divide theresult by 3.4 Move the decimal point 1 place to the left.Example:1 The 2-digit number chosen to multiply by 15 is 68. 2Multiply by 2:
50. 50. 2 x 68 = 120 + 16 = 136 3 Divide the result by 3: 136/3 = 45 1/34Move the decimal point 1 place to the left: 4.5 1/3 5 So 68divided by 15 = 4.51/3.See the pattern?1 The 2-digit number chosen to multiply by 15 is 96. 2Multiplyby 2: 2 x 96 = 180 + 12 = 192 3 Divide the result by 3: 192/3 =644 Move the decimal point 1 place to the left: 6.4 5 So 96/15 =6.4.With this method you will be able to divide numbers by 15with two quick1 Select a 2-digit number. (Choose larger numbers when youfeel sureabout the method.) 2 Multiply by 4 (or by 2 twice). 3 Move thedecimall point two places to the left.Example:1 The 2-digit number chosen to divide by 25 is 38. 2 Multiplyby 4: 4 x 38= 4 x 30 = 120 + 32 = 152. 3 Move the decimal point 2 placesto the left:1.52 4 So 38 divided by 25 = 1.52.See the pattern?1 The 3-digit number chosen to divide by 25 is 641. 2 Multiplyby 2 twice:2 x 64 = 1282. 2 x 1282 = 2400 + 164 = 2564.3 Move the decimal point 2 places to the left: 25.64. 4 So 641divided by25 = 25.64.1 Select a 2-digit number (progress to larger ones). 2 Multiplyit by 3. 3Move the decimal point 2 places to the left.
51. 51. Example:1 The 2-digit number chosen to multiply by 33 1/3 is 46. 2Multiply by 3:3 x 46 = 3(40 + 6) = 120 + 18 = 138 3 Move the decimal point 2places tothe left: 1.38 4 So 46 divided by 33 1/3 = 1.38. (If you divide by33.3using a calculator, you will not get the exact answer.)See the pattern?1 If the 3-digit number chosen to multiply by 33 1/3 is 650: 2Multiplyby 3: 3 x (600 + 50) = 1800 + 150 = 1950 3 Move the decimalpoint 2places to the left: 19.50 4 So 650 divided by 33 1/3 = 19.5.Practice multiplying left to right and this procedure willbecome aneasy one - and you will get exact answers, too.Dividing a 2- or 3-digit number by 351 Select a 2-digit number. (Choose larger numbers when youfeel sureabout the method.) 2 Multiply by 2. 3 Divide the resultingnumber by 7.4 Move the decimal point 1 place to the left.Example:If the number chosen to divide by 35 is 61: Multiply by 2: 2 x61 = 122.Divide by 7: 122/7 = 17 3/7 Move the decimal point 1 place tothe left:1.7 3/7 So 61 divided by 35 = 1.7 3/7.See the pattern?1 If the number chosen to divide by 35 is 44: 2 Multiply by 2:2 x 44 =88 3 Divide by 7: 88/7 = 12 4/7 4 Move the decimal point 1place to theleft: 1.2 4/7 5 So 44 divided by 35 = 1.2 4/7.
52. 52. Division done by calculator will give repeating decimals(unless theoriginal number is a multiple of 7), truncated by the limits ofthe display.The exact answer must be expressed as a mixed number.1 Select a 2-digit number. 2 Multiply by 8. 3 Divide theproduct by 3. 4Example:1 The 2-digit number chosen to divide by 37 1/2 is 32. 2Multiply by 8:8 x 32 = 240 + 16 = 256 3 Divide by 3: 256/3 = 85 1/3 4 Movethe decimalpoint two places to the left: .85 1/3 5 So 32 divided by 37 1/2 =.85 1/3.See the pattern?1 The 2-digit number chosen to divide by 37 1/2 is 51. 2Multiply by 8:8 x 51 = 408 3 Divide by 3: 408/3 = 136 4 Move the decimalpoint twoplaces to the left: 1.36 5 So 51 divided by 37 1/2 = 1.36.Dividing a 2-digit number by 451 Select a 2-digit number. 2 Divide by 5. 3 Divide the resultingnumberby 9.Example:1 f the number chosen to divide by 45 is 32: 2 Divide by 5:32/5 = 6.4 3 Divide the result by 9: 6.4/9 = .71 1/9 4 So 32dividedby 45 = .71 1/9.See the pattern?1 If the number chosen to divide by 45 is 61: 2 Divide by 5:61/5 =12.2 3 Divide the result by 9: 12.2/9 = 1.35 5/9 4 So 61 dividedby 451.35 5/9.
53. 53. 1 Select a 2-digit number. (Choose larger numbers when youfeelsure about the method.) 2 Multiply by 4 (or by 2 twice). 3Move thedecimal point two places to the left. 4 Divide by 3 (expressremainderas a fraction).Example:1 The 2-digit number chosen to divide by 75 is 82. 2 Multiplyby 4:4 x 82 = 328. 3 Move the decimal point 2 places to the left:3.28 4Divide by 3: 3.28/3 = 1.09 1/3 5 So 82 divided by 75 = 1.09 1/3.See the pattern?1 The 3-digit number chosen to divide by 75 is 631. 2 Multiplyby 4(multiply left to right): 4 x 631 = 2400 + 120 + 4 = 2520 + 4 =2524.3 Move the decimal point 2 places to the left: 25.24. 4 Divideby 3:25.24/3 = 8.41 1/3 5 So 631 divided by 75 = 8.41 1/3.Select a number.2 Multiply it by 2 3 Divide the result by 3.Example:1 The number chosen to divide by 1 1/2 is 72. 2 Multiply by 2:2 x 72 =144 3 Divide by 3: 144 / 3 = 48 4 So 72 divided by 1 1/2 = 48.See the pattern?1 The number chosen to divide by 1 1/2 is 83. 2 Multiply by 2:2 x 83 =166 3 Divide by 3: 166 / 3 = 55 1/3 4 So 83 divided by 1 1/2 =55 1/3.1 Select a 2-digit number. 2 Multiply by 3. 3 Divide by 4.Example:
54. 54. 1 The 2-digit number chosen to divide by 1 1/3 is 47. 2Multiply by 3:3 x 47 = 120 + 21 = 141 3 Divide by 4: 141/4 = 35 1/4 4 So 47dividedby 1 1/3 = 35 1/4.See the pattern?1 The 2-digit number chosen to divide by 1 1/3 is 82. 2Multiply by 3:3 x 82 = 246 3 Divide by 4: 246/4 = 61 1/2 4 So 82 divided by1 1/3 = 61 1/2.With this pattern you will be able to give answers quickly, butmostimportantly, your answers will be exact. If a calculator userdividesby 1.3, the answer will NOT be correct.1 Select a 2-digit number. 2 Multiply by 8. 3 Move the decimalpointone place to the left.Example:1 The 2-digit number chosen to divide by 1 1/4 is 32. 2Multiply by 8:8 x 32 = 240 + 16 = 256 3 Move the decimal point one place tothe left: 25.6 4 So 32 divided by 1 1/4 = 25.6.See the pattern?1 The 2-digit number chosen to divide by 1 1/4 is 64. 2Multiply by 8:8 x 64 = 480 + 32 = 512 3 Move the decimal point one place tothe left:51.2 4 So 64 divided by 1 1/4 = 51.2.Multiply from left to right for ease and accuracy. You willsoon bedoing this division by a mixed number quickly.1 Select a number. 2 Multiply it by 5 3 Divide the result by 6.Example:
55. 55. 1 The number chosen to divide by 1 1/5 is 24. 2 Multiply by 5:5 x 24= 120 3 Divide by 6: 120/6 = 20 4 So 24 divided by 1 1/5 = 20.See the pattern?1 The number chosen to divide by 1 1/5 is 76. 2 Multiply by 5:5 x 76= 350 + 30 = 380 3 Divide by 6: 380/6 = 63 2/6 4 So 76 dividedby 1 1/5= 63 1/3.Dividing a 2-digit number by 1 2/31 Select a 2-digit number. 2 Multiply by 6 (or by 2 and 3). 3Move the decimal point one place to the left.Example:1 The 2-digit number chosen to divide by 1 2/3 is 78. 2Multiplyby 3: 3 x 78 = 210 + 24 = 234 3 Multiply by 2: 2 x 234 = 468 4Movethe decimal point one place to the left: 46.8 5 So 78 dividedby1 2/3 = 46.8.See the pattern?1 The 2-digit number chosen to divide by 1 2/3 is 32. 2Multiply by 3:3 x 32 = 96 3 Multiply by 2: 2 x 96 = 180 + 12 = 192 4 Move thedecimalpoint one place to the left: 19.2 5 So 32 divided by 1 2/3 =19.2.Practice multiplying from left to right and you will becomeadept atmentally dividing a number by 1 2/3.1 Select a 2-digit number. 2 Multiply by 8. 3 Move the decimalpoint3 places to the left.Example:
56. 56. 1 The 2-digit number chosen to divide by 125 is 72. 2 Multiplyby 8:8 x 72 = 560 + 16 = 576. 3 Move the decimal point 3 places tothe left: .576 4 So 72 divided by 125 = .576.See the pattern?1 The 2-digit number chosen to divide by 125 is 42. 2 Multiplyby 8:8 x 42 = 320 + 16 = 336.3 ove the decimal point 3 places to the left: .336 4 So 42divided by125 = .336.1 Select a 2-digit number. 2 Multiply by 4. 3 Divide theproduct by 7.Example:1 The 2-digit number chosen to divide by 1 3/4 is 34. 2Multiply by 4:4 x 34 = 136 3 Divide the product by 7: 137/6 = 19 3/7 (If youuse acalculator, you will get a long, inexact decimal number.)4 So 34 divided by 1 3/4 = 19 3/7.See the pattern?1 The 2-digit number chosen to divide by 1 3/4 is 56. 2Multiply by 4:4 x 56 = 224 3 Divide the product by 7: 224/7 = 32 4 So 56dividedby 1 3/4 = 32.Notice that numbers divisible by 7 will produce whole-numberquotients. For numbers not divisible by 7, your calculator willgiveyou long decimal results that are not exact.1 Select a 2-digit number. 2 Multiply it by 7. 3 Move thedecimalpoint 1 place to the left.
57. 57. Example:1 The number chosen to multiply by 1 3/7 is 36. 2 Multiply by7:7 x 36 = 210 + 42 = 252 3 Move the decimal point 1 place totheleft: 25.2 4 So 36 divided by 1 3/7 = 25.2.See the pattern?1 The number chosen to multiply by 1 3/7 is 51. 2 Multiply by7:7 x 51 = 357 3 Move the decimal point 1 place to the left: 35.74So 36 divided by 1 3/7 = 35.7.1 Select a 2- or 3-digit number. 2 Multiply by 4 (or by 2 twice).3 Move the decimal point one place to the left.Example:1 The 2-digit number chosen to divide by 2 1/2 is 86. 2Multiplyby 4: 4 x 80 + 4 x 6 = 320 + 24 = 344 3 Move the decimal pointone place to the left: 34.4 4 So 86 divided by 2 1/2 = 34.4.See the pattern?1 The 3-digit number chosen to divide by 2 1/2 is 624. 2Multiplyby 2: 2 x 624 = 1248 3 Multiply by 2: 2 x 1248 = 2400 + 96 =24964 Move the decimal point one place to the left: 249.6 5 So 624divided by 2 1/2 = 249.6.Multiply by 4 when this is easy; otherwise use two steps andmultiply by 2 twice.Dividing a 2-digit number by 2 1/31 Select a 2-digit number. 2 Multiply by 3. 3 Divide the resultby 7.Example:1 he 2-digit number chosen to divide by 2 1/3 is 42. 2 Multiplyby 3:
58. 58. 3 x 42 = 126 3 Divide by 7: 126/7 = 18 4 So 42 divided by 2 1/3= 18.See the pattern?1 The 2-digit number chosen to divide by 2 1/3 is 73. 2Multiply by 3:3 x 73 = 219 3 Divide by 7: 219/7 = 31 2/7 4 So 73 divided by 21/3 = 31 2/7.If the number chosen is divisible by 7, the quotient will be awholenumber. If the number is not divisible by 7, a calculator userwillget a long, inexact decimal, while your answer will be exact.1 Select a 2-digit number. 2 Multiply by 3. 3 Divide by 8.Example:1 The 2-digit number chosen to divide by 2 2/3 is 32. 2Multiply by 3:3 x 32 = 96 3 Divide by 8: 96/8 = 12 4 So 32 divided by 2 2/3 =12.See the pattern?1 The 2-digit number chosen to divide by 2 2/3 is 61. 2Multiply by 3:3 x 61 = 183 3 Divide by 8: 183/8 = 22 7/8 4 So 61 divided by 22/3 = 22 7/8.Using this method, your answers will be exact. Those usingcalculators will only get approximations.1 Select a 2-digit number. 2 Mltmultiiply the number by 2. 3Divide theproduct by 7.Example:1 The 2-digit number chosen to divide by 3 1/2 is 42. 2Multiply by 2: 2 x42 = 84 3 Divide by 7: 84/7 = 12 4 So 42 divided by 3 1/2 = 12.See the pattern?1 The 2-digit number chosen to divide by 3 1/2 is 61. 2Multiply by 2:
59. 59. 2 x 61 = 122 3 Divide by 7: 122/7 = 17 3/7 4 So 61 divided by 31/2 = 17 3/7.If the number chosen is divisible by 7, the answer will be awhole number.For numbers not divisible by 7, a calculator will get arepeating decimal,but your fractional answer will be exact.1 Select a 2- or 3-digit number. 2 Multiply by 3. 3 Move thedecimal pointone place to the left.Example:1 The 2-digit number chosen to divide by 3 1/3 is 72. 2Multiply by 3:72 x 3 = 216 3 Move the decimal point one place to the left:21.6 4 So72 divided by 3 1/3 = 21.6.See the pattern?1 The 2-digit number chosen to divide by 3 1/3 is 48. 2Multiply by 3:48 x 3 = 120 + 24 = 144 3 Move the decimal point one place tothe left:14.4 4 So 48 divided by 3 1/3 = 14.4.After practicing, choose larger numbers. Insist on exactanswers (youwont get an exact answer if you divide by 3.3 using acalculator).Multiply from left to right in steps and impress your friendswithyour mental powers.Dividing a 2-digit number by 3751 Select a 2-digit number. 2 Multiply by 8. 3 Divide theproduct by3 (express remainder as a fraction). 4 Move the decimal pointthree places to the left.Example:
60. 60. 1 The number chosen to divide by 375 is 32. 2 Multiply by 8:8 x 32 = 240 + 16 = 256 3 Divide by 3: 256/3 = 85.3 1/3 4 Movethe decimal point 3 places to the left: .0853 1/3 5 So 32divided by 375 = .0853 1/3.See the pattern?1 The number chosen to divide by 375 is 61. 2 Multiply by 8:8 x 61 = 480 + 8 = 488 3 Divide by 3: 488/3 = 162 2/3 4 Movethe decimal point 3 places to the left: .162 2/3 5 So 61 dividedby 375 = .162 2/3.72 3 Divide by 9:72/9 = 8 4 So 36divided by 4 1/2 =8.For numbers notdivisible by 9, yourcalculator will getarepeating decimal,but your fractionalanswer will beexact.Dividing a 2-digit number by6251 Select a 2-digit number. 2 Multiply by 8. 3 Divide theproductby 5. 4 Move the decimal point 3 places to the left.Example:1 The 2-digit number chosen to divide by 625 is 65. 2 Multiplyby8: 8 x 65 = 480 + 40 = 520 3 Divide by 5: 520/5 = 104 4 Movethedecimal point 3 places to the left: .104 5 So 65 divided by 625= .104.See the pattern?
61. 61. 1 The 2-digit number chosen to divide by 625 is 32. 2 Multiplyby8: 8 x 32 = 240 + 16 = 256 3 Divide by 5: 256/5 = 51.2 4 Movethedecimal point 3 places to the left: .0512 5 So 32 divided by625 =.0512.Dividing a 2-digit number by 7 1/21 Select a 2-digit number. 2 Multiply it by 4 (or by 2 twice). 3Divideby 3. 4 Move the decimal point 1 place to the left.2M u lt i p lExample: y b 1 The 2-digit number chosen to multiply by 71/2 is 42. y4: 42 x 4 = 168 3 Divide by 3: 168/3 = 56 4 Move the decimalpoint1 place to the left: 5.6 5 So 42 divided by 7 1/2 = 5.6.See the pattern?1 The 2-digit number chosen to multiply by 7 1/2 is 93. 2Multiply byDividing a 2-digit number by 4 1/221 Select a 2-digit number.2 Multiply the number by 2.3 Divide the product by 9.Example:1 The 2-digit number chosen to divide by 4 1/2 is 62.23Multiply by 2: 2 x 62 = 124 Divide by 9: 124/9 = 13 7/94 So 62 divideby 4 1/2 = 13 7/9.See the pattern?1 The 2-digit number chosen to divide by 4 1/2 is 36.4: 93 x 4 = 360 + 12 = 372 3 Divide by 3: 372/3 = 124 4 Movethe
62. 62. decimal point 1 place to the left: 12.4 5 So 93 divided by 7 1/2= 12.4.Practice and you will soon be cranking out these quotientswithspeed and accuracy.Dividing a 2- or 3-digit number by 16 2/31 Select a 2-digit number. (Choose larger numbers when youfeelsure about the method.) 2 Multiply by 6 (or by 3 and then 2). 3Move the decimal point two places to the left.Example:1 The 2-digit number chosen to divide by 16 2/3 is 72. 2Multiplyby 3: 3 x 72 = 216 3 Multiply by 2: 2 x 216 = 432 4 Move thedecimalpoint 2 places to the left: 4.32 5 So 72 divided by 16 2/3 =4.32.See the pattern?1 The 2-digit number chosen to divide by 16 2/3 is 212. 2Multiply by3: 3 x 212 = 636 3 Multiply by 2: 2 x 636 = 1200 + 72 = 1272 4Move thedecimal point 2 places to the left: 12.72 5 So 212 divided by16 2/3 = 12.72.Practice multiplying by 3, then by 2, and you will be able todo theseproblems quickly.Divisibility RulesWhy do these rules work? - Dr. RobDivisibilidad por 13 y por números primos (13,17,19...)-en español, de la lista SNARKFrom the Archives of the Math Forums Internet project AskDr. Mathourthanks to Ethan Dr. Math Magness, Steven Dr. MathSinnott,
63. 63. nd, for the explanation of why these rules work, Robert L.Ward (Dr. Rob).Dividing by 3Add up the digits: if the sum is divisible by three, then thenumber isas well. Examples: 1 111111: the digits add to 6 so the wholenumberis divisible by three. 2 87687687. The digits add up to 57, and5 plusseven is 12, so the original number is divisible by three.Why does the divisibility by 3 rule work?From: "Dr. Math" To: keving@ecentral.com (Kevin Gallagher)Subject:Re: Divisibility of a number by 3As Kevin Gallagher wrote to Dr. MathOn 5/11/96 at 21:35:40 (Eastern Time),>Im looking for a SIMPLE way to explain to several verybright 2nd>graders why the divisibility by 3 rule works, i.e. add up allthe >digits;if the sum is evenly divisible by 3, then the number is as well.>Thanks!>Kevin GallagherThe only way that I can think of to explain this would be asfollows:Look at a 2 digit number: 10a+b=9a+(a+b). We know that 9a isdivisible by 3, so 10a+b will be divisible by 3 if and only if a+bis.Similarly, 100a+10b+c=99a+9b+(a+b+c), and 99a+9b isdivisibleby 3, so the total will be iff a+b+c is.This explanation also works to prove the divisibility by 9 test.Itclearly originates from modular arithmetic ideas, and Im notsure
64. 64. if its simple enough, but its the only explanation I can thinkof.Doctor Darren, The Math ForumCheck out our web site -http://forum.swarthmore.edu/dr.math/Dividing by 4ook at the last two digits. If they are divisible by 4, thenumber isas well. Examples:1 100 is divisible by 4. 2 1732782989264864826421834612 isdivisible by four also, because 12 is divisible by four.Dividing by 5If the last digit is a five or a zero, then the number is divisibleby 5.Dividing by 6Check 3 and 2. If the number is divisible by both 3 and 2, it isdivisible by 6 as well. Robert Rusher writes in:Another easy way to tell if a [multi-digit] number isdivisible by six . . .is to look at its [ones digit]: if it is even, and the sum ofthe [digits] isa multiple of 3, then the number is divisible by 6.Dividing by 7To find out if a number is divisible by seven, take the lastdigit,double it, and subtract it from the rest of the number.Example:If you had 203, you would double the last digit to get six, andubtract that from 20 to get 14. If you get an answer divisibleby 7(including zero), then the original number is divisible byseven.If you dont know the new numbers divisibility, you canapplythe rule again.Dividing by 8
65. 65. Check the last three digits. Since 1000 is divisible by 8, if thelast three digits of a number are divisible by 8, then so is thewhole number. Example: 33333888is divisible by 8; 33333886isnt. How can you tell whether the last three digits aredivisible by 8? Phillip McReynolds answers:If the first digit is even, the number is divisible by 8 if thelast two digits are. If the first digit is odd, subtract 4 fromthe last two digits; the number will be divisible by 8 if the resulting last two digits are. So, to continue the last example,33333888 is divisible by 8 because the digit in the hundredsplace is an even number, and the last two digits are 88, whichis divisible by 8. 33333886 is not divisible by 8 because thedigit in the hundreds place is an even number, but the lasttwo digits are 86, which is not divisible by 8.Dividing by 9Add the digits. If they are divisible by nine, then the numberis as well. This holds for any power of three.Dividing by 10If the number ends in 0, it is divisible by 10.Dividing by 11Lets look at 352, which is divisible by 11;the answer is 32. 3+2 is 5; another way to say thisis that 35 -2 is 33. Now look at 3531, which is alsodivisible by 11. It is not a coincidence that 353-1 is352 and 11 x 321 is 3531. Here is a generalizationof this system. Lets look at the number 94186565.First we want to find whether it is divisible by 11, but on theway we are going to save the numbers that we use: in everystep we will subtract the last digit from the other digits, thensave the subtracted amount in order. Start with9418656 - 5 = 9418651 SAVE 5 Then 941865 - 1 = 941864SAVE 1Then94186 - 4 = 94182 SAVE 4 Then 9418 - 2 = 9416 SAVE 2 Then941- 6 = 935
66. 66. SAVE 6 Then 93 - 5 = 88 SAVE 5 Then 8 - 8 = 0 SAVE 8Now write the numbers we saved in reverse order, and wehave8562415, which multiplied by 11 is 94186565.Heres an even easier method, contributed by ChisForen:Takeany number, such as 365167484.Add the 1,3,5,7,..,digits.....3 + 5 + 6 + 4 + 4 = 22Add the 2,4,6,8,..,digits.....6 + 1 + 7 + 8 =22If the difference, including 0, is divisible by 11, then so isthenumber. 22 - 22 = 0 so 365167484 is evenly divisible by11.See alsoDivisibility by 11 in the Dr. Math archives.Dividing by 12Check for divisibility by 3 and 4.Dividing by 13Heres a straightforward method supplied by Scott Fellows:D e lete the last digit from the given number. Then subtractninetimes the deleted digit from the remaining number. If what isleftis divisible by 13, then so is the original number.And heres a more complex method that can be extended tootherformulas:1 = 1 (mod 13)10 = -3 (mod 13) (i.e., 10 - -3 is divisible by13)100= -4 (mod 13) (i.e., 100 - -4 is divisible by 13)1000 = -1 (mod13)(i.e., 1000 - -1 is divisible by 13)10000 = 3 (mod 13)100000 = 4(mod 13)1000000 = 1 (mod 13)Call the ones digit a, the tens digit b, the hundreds digit c, .....and you get:a - 3*b - 4*c - d + 3*e + 4*f + g - .....
67. 67. If this number is divisible by 13, then so is the originalnumber.You can keep using this technique to get other formulas fordivisibility for prime numbers. For composite numbers justcheck for divisibility by divisors.Finding 2 1/2 percent of a number1 Choose a number (start with 2 digits and advance to 3 withpractice).2 Divide by 4 (or divide twice by 2). 3 Move the decimal pointone placeto the left.Example:1 If the number selected is 86: 2 Divide 86 by 4: 86/4 = 21.5 3Move thedecimal point one place to the left.: 2.15 4 So 2 1/2% of 86 =2.15.See the pattern?1 If the number selected is 648: 2 Divide 648 by 2 twice: 648/2= 324,324/2 = 162 3 Move the decimal point one place to the left.:16.2 4 So2 1/2% of 648 = 16.2.Practice dividing by 4, or by 2 twice, and you will be able tofind theseanswers faster than with a calculator.Finding 5 percent of a number1 Choose a large number (or sum of money). 2 Move thedecimalpoint one place to the left. 3 Divide by 2 (take half of it).Example:1 If the amount of money selected is \$850: 2 Move thedecimal pointone place to the left.: 85 3 Divide by 2: 85/2 = 42.50 4 So 5%of \$850 = \$42.50.See the pattern?
68. 68. 1 If the amount of money selected is \$4500: 2 Move thedecimal pointone place to the left.: 450 3 Divide by 2: 450/2 = 225 4 So 5%of \$4500 =\$225.Finding 15 percent of a number1 Choose a 2-digit number. 2 Multiply the number by 3. 3Divide by 2. 4 Move the decimal point one place to the left.Example:1 If the number selected is 43: 2 Multiply by 3: 3 x 43 = 129 3Divideby 2: 129/2 = 64.5 4 Move the decimal point one place to theleft: 6.455 So 15% of 43 = 6.45.See the pattern?1 If the number selected is 72: 2 Multiply by 3: 3 x 72 = 216 3Divide by2: 216/2 = 108 4 Move the decimal point one place to the left:10.85S o15%of72=10.8.Finding 45 percent of a number1 Choose a 2-digit number. 2 Multiply the number by 9. 3Divide by 2.4 Move the decimal point one place to the left.Example:1 If the number selected is 36: 2 Multiply by 9: 9 x 36 = 270 +54 = 324 3Divide by 2: 324/2 = 163 4 Move the decimal point one placeto the left: 16.2 5 So 45% of 36 = 16.2.See the pattern?
69. 69. If the number selected is 52: Multiply by 9: 9 x 52 = 450 + 18 =468 Divideby 2: 468/2 = 234 Move the decimal point one place to theleft: 23.4 So45% of 52 = 23.4.Finding 55 percent of a numbern 1 Choose a 2-digit number. e 2 Multiply the number by 11.(Add digitsfrom right to left - see examples). x 3 Divide by 2. t 4 Move thedecimalpoint one place to the left. di Example: git 1 If the number selected is 81: t 2 Multiply by 11: 11 x 81 =891 o rightdigit is 1 l eft is 1 + 8 = 9 last digit to left is 83 Divide by 2: 891/2 = 445.5 4 Move the decimal point oneplace to theleft: 44.55 5 So 55% of 81 = 44.55.See the pattern?1 If the number selected is 59:2. Multiply by 11: 11 x 59 = 649 right digit is 9 next digit to leftis 9 + 5= 14 (use the 4 and carry 1)last digit to left is 5 + 1 = 62 Divide by 2: 649/2 = 324.53 Move the decimal point one place to the left: 32.454 So 55% of 59 = 32.45.Why do the divisibility "rules" work?The Power of Modulo ArithmeticModulo arithmetic is a powerful tool that can be used to testfordivisibility by any number. The great disadvantage of usingmodulo arithmetic to test for divisibility is the fact that it isusuallya slow method. Under some circumstances, however, the
70. 70. application of modulo arithmetic leads to divisibility rulesthat canbe used. These rules are important to mathematics becausetheysave us lots of time and effort.In fact, many of the divisibility rules that are commonly used(rules for 3, 9, 11) have their roots in modulo arithmetic. Thispage will show you how some of these common divisibilityrules are connected to modulo arithmetic.Applying the Rules of Modulo ArithmeticThe rules of modulo arithmetic state that the number N isdivisibleby some number (P) if the above expression is also divisibleby Pafter the base (10) is replaced by the remainder of 10 dividedby P.In compact notation, this remainder is denoted by (10 modP). Thus, N is divisible by P if the following expression is alsodivisible by P:[Dn * (10 mod P)^(n-1)] + ... + [D2 * (10 mod P)^1] + [D1 * (10mod P)^0].Origin of Divisibility Rules for 3, 9, and 11D We can now see how the divisibility rules for P = 3, 9, and11 arerooted in modulo arithmetic. nConsider the case where P = 3. .Because 10 mod 3 is equal to 1, any number is divisible by 3if thefollowing expression is. also divisible by 3: .-[Dn * (1)^n-1] + ... + [D2 * (1)^1] + [D1* D2(1)^0]. Because 1 raised to any power is equal to 1, the above
71. 71. +expression can be simplified as: Dn + ... + D2 + D1 + D0,Dwhich is equal to the sum of digits in the number. Thusmoduloarithmetic allows us to 1state that any number is divisible by 3 if the sum of its digits isBecause 10 mod 9 is also 1, any number is divisible by 9 isthesum of its digits are alsodivisible by 9. Thus, the divisibility rules for 3 and 9 comedirectlyfrom modulo narithmetic.n)Finally, consider the case where P = 11. aBecause 10 mod 11 is equal to -1, any number is divisible by11if the n following expression is also divisible by 11: dD[Dn * (-1)^n-1] + ... + [D2 * (-1)^1] + [D1* (-1)^0].nBecause -1 raised to any even power is equal to 1 and -1raised toany odd +power is equal to -1, the above expression becomesa summation of the digits involving alternated signs: .. . - D2 + D1 (for odd n).This "alternating" summation rule for P = 11 is well knownand has two versions. The first version states a number isdivisible by 11 if the alternating sum of the digits is alsodivisilbeby 11 (i.e., the alternating sum is a muliple of 11). The secondversion states that a number is divisible by 11 if the sum ofeveryother digit starting with the rightmost digit is equal to thesum of
72. 72. every other digit starting with the second digit from the right(i.e.,the alternating sum is 0). Both of these versions comedirectlyfrom the rules of modulo arithmetic.Bases Other Than 10v Up to this point, we have only considered numbers writteninbase 10. A number can, however, be i written in any base (B).Such a number can be expressed by the followingsummation: git N = [Dn * B^n-1] + [Dn-1 * B^(n-2)] + ... + [D2 * B^1] + [D1*B^0]. s"The rules of modulo arithmetic apply no matter what base isused. These rules tell us that N is divisible by P if the followingexpression is also divisible by P: [Dn * (B mod P)^n-1] + ...+ [D2 * (B mod P)^1] + [D1* (B mod P)^0]. Most bases otherthan 10 are difficult to use, but basesthat are powers of 10 can easily beused to construct divisibility rules using modulo arithmetic.Consider writing the base-10 number N = 1233457 in base-100.This can be done by starting with the rightmost digit andgrouping the digits in pairs of two. bEach grouping oftwo digits is considered a single "digit" when the numberis ywritten in base-100. Written in base-100, the number is1 23 34 57 where 57, 34, 123, and 1 are considered single"digits." In a similar fashion, this number written 1inbase-1000 is 1 233 457 where 457, 233, and 1 are consideredsingle "digits." .Once the base-10 digits are grouped to formthe "digits," the above expression Bcan be used to test fordivisibility. Consider P = 7.Use the base, B = 1000. Because 1000 mod 7 = -1, the