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Number Systems Theory for CAT 2009 Quant

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Narrative about Number systems in Quantitative Aptitude for CAT

Narrative about Number systems in Quantitative Aptitude for CAT


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  • Very good article. Can you please send to gentleman208@gmail.com
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  • Number Systems Theory for CAT 2009 Quant
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  • Very Helpful!!! Thank you for posting
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  • awesome... very useful... thanks nickyswetha
    can u send me to my mail id : naveenkumarrs37@gmail.com
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  • 1. NUMBER SYSTEMS http://nov15.wordpress.com/ Presents QUANT For CAT 2009
  • 2. NUMBER SYSTEMS
  • 3. INTRODUCTION
  • 4. PRIME NUMBERS A number is prime if it is not divisible by  any prime number less than it’s square root. Ex: Is 179 a prime number ?  179 13.3 Prime Numbers less than 13.3 are  2,3,5,7,11,13 179 is not divisible by any of them, 179 is  prime.
  • 5. DIVISIBILITY RULES
  • 6. SOME MORE DIVISIBILITY RULES Test for divisibility by 7: Double the last digit and subtract it  from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82.  Now 82-12=70. This is divisible by 7, so 826 is divisible by 7. Test for divisibility by 11: Subtract the last digit from the  remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary. Example: 19151 --> 1915-1 =1914 –>191-4=187 –>18-7=11, so  yes, 19151 is divisible by 11.
  • 7. SOME MORE DIVISIBILITY RULES Test for divisibility by 13: Add four times the last digit to the remaining leading  truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary. Example: 50661–>5066+4=5070–>507+0=507–>50+28=78 and 78 is 6*13, so  50661 is divisible by 13. Test for divisibility by 17: Subtract five times the last digit from the remaining  leading truncated number. If the result is divisible by 17, then so was the first number. Apply this rule over and over again as necessary. Example: 3978–>397-5*8=357–>35-5*7=0. So 3978 is divisible by 17.  Test for divisibility by 19: Add two times the last digit to the remaining leading  truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary. Example:101156–>10115+2*6=10127–>1012+2*7=1026–>102+2*6=114 and  114=6*19, so 101156 is divisible by 19.
  • 8. REMEMBER IT! Here is a table using which you can easily remember the previous divisibility rules. Read the table as follows : For divisibility by 7 , subtract 2 times the last digit with the truncated number. Divisibility by Test 7 Subtract 2 x Last digit 11 Subtract 1 x Last digit 13 Add 4 x Last Digit 17 Subtract 5 x Last Digit 19 Add 2 x Last Digit
  • 9. UNIT’S DIGIT OF A NUMBER Find the unit’s digit of 71999 (7 to the power 1999)  Step 1: Divide the exponent by 4 and note down remainder  1999/4 => Rem = 3 Step 2: Raise the unit’s digit of the base (7) to the  remainder obtained (3) 73 = 343 Step 3: The unit’s digit of the obtained number is the required  answer. 343 => Ans 3 If the remainder is 0, then the unit’s digit of the base is raised to 4 and  the unit’s digit of the obtained value is the required answer. If Rem = 0 , Then 74 = XX1 -> Ans 1 Note: For bases with unit’s digits as 1,0,5,6 the unit’s digit for any  power will be the 1,0,5,6 itself. 453
  • 10. LAST TWO DIGITS OF A NUMBER We will discuss the last two digits of numbers  ending with the following digits in sets : a) 1  b) 3,7 & 9  c) 2, 4, 6 & 8 
  • 11. LAST TWO DIGITS OF A NUMBER a) Number ending with 1 :   Ex : Find the last 2 digits of 31786  Now, multiply the 10s digit of the number with the last digit of exponent 31786 = 3 * 6 = 18 -> 8 is the 10s digit. Units digit is obviously 1  So, last 2 digits are => 81 
  • 12. LAST TWO DIGITS OF A NUMBER b) Number Ending with 3, 7 & 9   Ex: Find last 2 digits of 19266 We need to get this in such as way that the  base has last digit as 1 19266 = (192)133 = 361133 Now, follow the previous method => 6 * 3 =  18  So, last two digits are => 81
  • 13. LAST TWO DIGITS OF A NUMBER b) Number Ending with 3, 7 & 9  Remember :   34 = 81  74 = 2401  92 = 81
  • 14. LAST TWO DIGITS OF A NUMBER Ex 2: Find last two digits of 33288  Now, 33288 = (334)72 = (xx21)72 Ten’s digit is -> 2*2 = 04 -> 4 So, last two digits are => 41 Ex 3: find last 2 digits of 87^474  (872)*(874)118 => (xx69) * (xx61)118 (6 x 8 = 48) => (xx69)*(81) So, last two digits are 89
  • 15. LAST TWO DIGITS OF A NUMBER c)Ending with 2, 4, 6 or 8  Here, we use the fact that 76 power any number gives 76. We also need to remember that,  242 = xx76  210 = xx24  24even = xx76  24odd = xx24
  • 16. LAST TWO DIGITS OF A NUMBER Ex: Find the last two digits of 2543 2543 = ((210)54) * (23) = ((xx24)54)* 8 = ((xx76)27)*8 76 power any number is 76 Which gives last digits as => 76 * 8 = 608   So last two digits are : 08
  • 17. HIGHEST POWER Highest power of a number that divides the  factorial of another number. What is the highest power of 5 that divides  60!(factorial) Note: N! = N*(N-1)*(N-2)*(N-3)….(2)*(1)  Now, Continuously divide 60 with 5 as shown  60/5 = 12, 12/5 = 2 (omit remainders) 2/5 = 0 <- stop at 0 Now add up all the quotients => 12+2+0 = 14  So highest power of 5 that divides 60! is 14. 
  • 18. HIGHEST POWER Ex: Find Highest power of 15 that divides 100!  Here, as 15 is not a prime number we first split 15 into prime factors.  15 = 5 * 3 Now, find out highest power of 5 that divides 100! and also highest power of 3  that divides 100! . For 5 : 100/5 =20  20/5 = 4 4/5 = 0 So, 20 + 4 + 0 = 24 For 3 : 100/3 = 33  33/3 = 11 11/3 = 3 3/3 = 1 1/3 = 0 So, 33 + 11+ 3 + 1 + 0 = 48 Now, the smallest number of these is taken which will be 24. 
  • 19. NUMBER OF ZEROES Ex: Find the number of zeroes in 75!   This means highest power of 10 which can divide 75! 10 = 5*2  If we consider highest power of 5 which can divide 75! , it’s sufficient. 75/5 =15 15/5 =3 3/5 =0 So, 15+3+0 = 18 So, there are 18 zeroes in 75! 
  • 20. NUMBER OF FACTORS OF A NUMBER If the number N can be expressed as a product of prime  factors such that N = (pa)*(qb)*(rc) where, p,q,r = prime factors a,b,c = powers to which each is raised Then,  No. of factors of N (including 1, N) = (a+1)*(b+1)*(c+1)*….
  • 21. EVEN AND ODD Even number => Divisible by 2   Odd Number => Not Divisible by 2  Important Results : exe=e exo=e oxo=o
  • 22. EXERCISE Download the related exercise here Exercise 1 - Number Systems
  • 23. LET ME KNOW!!! If you liked this presentation, do comment on http://nov15.wordpress.com or write to Nicky at nickyswetha20@yahoo.com
  • 24. THANK YOU!