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Target 320+ in GRE
1
GREKing
Vedic Numbers
 Number Tree
 LCM HCF
 Divisibility Rules
 Power cycle
 Remainder Theorem
 Remainder of powers an – bn
 Last and Second last digit
 Power of Exponents
 Euler’s Theorem
 Fermet’s Theory
 Wilson Theorem
 Number Systems (decimal binary)
Importance in exams
CAT – 4 questions
CMAT – 2 questions
NMAT – 5 questions
Number Tree
HCF & LCM
42 = 21 x 2
= 7 x 3 x 2
• Factorize the following numbers:
• 42
• 72
• 84
• 65
• 108
• 210
• Calculate the LCM and HCF
for the following groups of
numbers
a) 42, 70
b) 18, 24, 60 ,150
LCM of nos 60
• Three bells ring after 3, 4 and 5 minutes
respectively. If they start ringing together, when
• will they ring together again?.
a) 50
b) 90
c) 60
d) 42
HCF of nos 10
• Three buckets of milk having capacities 40, 30 and 50 L
respectively. What is the largest
• size of the measure that can be used to pour milk from all
three buckets such that the
• volumes of milk contained in each bucket is an integral
number of pourings used by the
• measure.
a) 50
b) 10
c) 60
d) 42
A = 12 x
B = 12 y
144 xy = 2880
xy = 20
1, 20
2, 10
4, 5
5,4
10,2
20,1
• The product of two numbers is 2880. If the
HCF is 12, how many such pairs of number are
possible?
a) 3
b) 6
c) 8
d) 2
LCM of nos
720
• Find the least number which when divided by
9,12,16,30 leaves in each case a
• remainder of 3.
a) 721
b) 722
c) 723
d) 754
LCM of nos 15
105
120
135…..
• Find the number of all the numbers divisible
by 3 and 5 between 100 to 200.
a) 9
b) 8
c) 6
d) 7
M(3) + M(5) – M(3,5)
30 + 20 – 7
= 43
• Find the number of all the numbers divisible by
3 or 5 between 100 to 200.
a) 25
b) 89
c) 60
d) 43
LCM of nos 15
105
120
135…..
Sn = n/2 (a+l)
• Sum of all the numbers divisible by 3 and 5
between 100 to 200.
a) 1050
b) 8090
c) 6020
d) 4231
120
• What is the least three digit number that is
divisible by 5 and 6?
• - What is the least 4 digit number that is
divisible by 12 and 15?
• - What is the least 4 digit number that is
divisible by 25 and 30?
Rules of Divisibility
TESTS OF DIVISIBILITY:
• Divisibility By 2 : A number is divisible by 2, if its unit's
digit is any of 0, 2, 4, 6, 8.
• Ex. 84932 is divisible by 2, while 65935 is not.
• Divisibility By 3 : A number is divisible by 3, if the sum
of its digits is divisible by 3.
• Ex.592482 is divisible by 3, since sum of its digits = (5 +
9 + 2 + 4 + 8 + 2) = 30, which is divisible by 3.
• But, 864329 is not divisible by 3, since sum of its digits
=(8 + 6 + 4 + 3 + 2 + 9) = 32, which is not divisible by 3.
Last 3 multiple of 8
• Find the value of K when 425K is divisible by 8..
a) 6
b) 5
c) 0
d) 4
Rule of 2 than 9
• A number A4571203B is divisible by 18. Which
of the
• following values can A and B take ?
a) 1, 2
b) 2, 3
c) 6, 8
d) 3, 3
Rule of 8 and 11
• Find A and B , if 3765A56682B is divisible by
88?
21 = 2
22 = 4
23 = 8
24 = 16
25 = 32
……
2
4
8
6 4k
Power Cycle
4k = 2, 3, 7, 8
2k = 4, 9
1k = 1, 5, 6, 0
• 1 = 1, 1, 1, …
• 2 = 2, 4, 8, 6…
• 3 = 3, 9, 7, 1…
• 4 = 4, 6, 4, 6
• 5 = 5, 5, 5, 5…
• 6 = 6, 6, 6, 6…
• 7 = 7, 9, 3, 1…
• 8 = 8, 4, 2, 6…
• 9 = 9, 1, 9, 1…
• 0 = 0, 0, 0, 0
• What is the last digit of 360
• What is the last digit of 743
• What is the last digit of 940
• What is the last digit of 56120
• What is the last digit of 26670
• Find the unit's digit in
• (264)102 + (264)103
a) 0
b) 2
c) 4
d) 20
NMAT• Find the unit's digit in
• (263)102 x (265)103
a) 5
b) 2
c) 4
d) 20
Remainder Theorem
• Rem (X / A) = Rem (Y/A) x Rem (Z/A)
• Where y and z are factors of X
• Find the remainder of
a) 24^3 / 7
b) 72^6 / 12
c) 625^12 / 12
• Find the remainder when 2^246 is divided by
7.
a) 1
b) 3
c) 7
d) 9
e) none of these
3^2720
• Find the right most non zero digit of (30)2720
• a.1 b.3 c.7 d.9.
Euler’s Theorem
N + 1 / N is always 1
N – 1 / N is always 1, -1
• Find the remainder when 3560 is divided by 2.
a) 5
b) 2
c) 4
d) 1
N + 1 / N is always 1
N – 1 / N is always 1, -1
• Find the remainder when 9560 is divided by 2 =
• 33560 is divided by 2 =
• 5560 is divided by 4 =
N – 1 / N is always 1, -1
• Find the remainder when 7560 is divided by 2 =
• 31560 is divided by 2 =
• 3560 is divided by 4 =
16 / 17 = -1
So 2 – 1 = 1
CAT 2002
• Find the remainder when 2256 is divided by 17 =
a) 3
b) 1
c) 7
d) 9.
2.2256
16 / 17 = -1
So 2 – 1 = 1
• Find the remainder when 2257 is divided by 17 =
A. 3
B. 2
C. 7
D. 9
Cyclic Remainders
541 / 7 = 5
542 / 7 = 25 / 7 = 4
543 / 7 = 20 / 7 = 6
544 / 7 = 2
545 / 7 = 3
546 / 7 = 1
XAT LEVEL
• Find the remainder when 54120 is divided by 7 =
a) 3
b) 1
c) 7
d) 9.
41 / 6 = 4
42 / 6 = 4
……
CAT 2003
• Find the remainder when 496 is divided by 6 =
a) 3
b) 4
c) 7
d) 9
•An+Bn
• Is always divisible by A + B for all
odd n
•An - Bn
• Is always divisible by A - B for all odd n
• Is always divisible by A - B for all even n
• Is always divisible by A + B for all even n
351 - 231 = 12
SNAP
• Find the remainder when
• 3523 - 2323 is divided by 12
a) 3
b) 0
c) 7
d) 9
352 - 232 = 12, 58
• Find the remainder when
• 3522 - 2322 is divided by 58
a) 3
b) 0
c) 7
d) 9.
7+6 = 13;
73 – 63 = 127
CAT 2002
• Find the remainder when
• 76n – 66n is divided by ?
a) 13
b) 127
c) 559
d) All of these
Highest powers in n!
780/7 + 780/49 + 780/343
CMAT
• Find the highest power of 7 which will divide 780!
a) 13
b) 127
c) 559
d) 15
250/5 + 250/25 + 250/125
• Find the highest power of
• 25 which will divide 250!
a) 13
b) 127
c) 559
d) 31
Wilson's Theorem
• Let p be an integer greater than one. p is prime if and only if (p-1)! = -1 (mod p).This
beautiful result is of mostly theoretical value because it is relatively difficult to
calculate (p-1)! In contrast it is easy to calculate ap-1, so elementary primality tests are
built using Fermat's Little Theorem rather than Wilson's. Neither Waring or Wilson
could prove the above theorem, but now it can be found in any elementary number
theory text. To save you some time we present a proof here.
• Proof. It is easy to check the result when p is 2 or 3, so let us assume p > 3. If p is
composite, then its positive divisors are among the integers1, 2, 3, 4, ... , p-1 and it is
clear that gcd((p-1)!,p) > 1, so we can not have (p-1)! = -1 (mod p).
• However if p is prime, then each of the above integers are relatively prime to p. So for
each of these integers a there is another such that ab = 1 (mod p). It is important to
note that this b is unique modulo p, and that since p is prime, a = b if and only if a is 1
or p-1. Now if we omit 1 and p-1, then the others can be grouped into pairs whose
product is one showing2.3.4.....(p-2) = 1 (mod p) (or more simply (p-2)! = 1 (mod p)).
Finally, multiply this equality by p-1 to complete the proof.
Vedic Numbers
Rem. Of(17/12*19/12*21/12)
• Let N=1421*1423*1425.What is the remainder
when N is divided by 12?
a) 0
b) 9
c) 3
d) 6
2^4+2/5=3.
• What is the remainder when (2^100 + 2) is
divided by 101?
a) 3
b) 4
c) 5
d) 6
2^n & 5^n=n+1.
• The values of numbers 2^2004 and 5^2004 are
written one after another. How many digits are
there in all?
a) 4008
b) 2003
c) 2005
d) none of these
3^2720=1.
• Find the right most non zero digit of
(30)^2720.
a) 1
b) 3
c) 7
d) 9.
A^n + B^n=divisible by a+b
• The remainder, when (15^23 + 23^23) is divided by
19 is:
a) 4
b) 0
c) 15
d) 18.
Divisibility rule of 27
• What is the remainder of 22222222222….(27
times) is divided by 27?
a) 0
b) 1
c) 3
d) 4
N^n+1 /n=1
• Find the remainder when 2009^2010 is divided
by 2011.
a) 2010
b) 0
c) 1
d) 2009
Understand the format of Q
• 111^111 +222^111 +333^111
+444^111+….999^111 is divided by 555?
a) 1
b) 0
c) 2
d) 3
e) 4
1! +2*2!/3=rem. is 2
• Let N!=1*2*3*4*5*6*N for N greater or equal to 1.
• If P=1! +(2*2!) +(3*3!) +(4*4!)……(12*12!). Find the
remainder when P is divided by 13.
a) 11
b) 1
c) 2
d) 12
• If R=30^65 -29^65/30^64+29^64. then,
a) 0<R<1
b) 0.5<R<1
c) 0<0.1
d) R>1
306 is divisible by 3, 6, 9
• 3066 - 306 is not divisible by which of the
following ?
a) 3
b) 4
c) 6
d) 9
We need 1 even and two odd
• P,Q& R form a set if distinct prime numbers less
than 20.How many such prime numbers are
possible for which the sum of P,Q & R is even.
a) 15
b) 21
c) 24
d) 28
• Q5. (153)8 = (x)10.
a) 162
b) 107
c) 166
d) 207
e) 203
Find the last two digits if 1733^148.
• Finding the last n second last digit
• (2^10)even= 24
• (2^10)odd =76
• Finding the last n second last digit
• For odd digits
• (21)^27= 41
• (31)^21=31
• (a1)^cd = ad1
Concept of perfect squares
examining the nature of all perfect
squares…
Any number will have a unit’s digit
as..
1,2,3,4,5,6,7,8,9,0.
Examining the nature of 1.
1^2=01
11^2=121
21^2=441
31^2=961 n so on…
Nature of 2.
2^2=04
12^2=144
22^2=484
32^2=1024 n so on…
Nature of 3.
3^2=09
13^2=169
23^2=529
33^2=1089..
Nature of 4
4^2=16
14^2=196
24^2=576
34^2=1156..
Nature of 5
5^2=25
15^2=225
25^2=625
35^2=1225..
Nature of 6
6^2=36
16^2=256
26^2=576..
Nature of 7
7^2=49
17^2=289
27^2=729..
Nature of 8
8^2=64
18^2=324
28^2=784..
Nature of 9
9^2=81
19^2=361
29^2=841..
74= 2401 => last digit 1
78= 5764801 => last digit
1
72008 = 1 , cyclist of four.
What is the last two digits of 72008
a) 71
b) 51
c) 01
d) 11
X can be even or odd & same
for Y.
If x2 - y2 = 1234
Find the number of integral solutions of x,y.
Where x, y < 150.
a) 4
b) 5
c) 0
d) 8
To get the last digit as 9,the
only ways are
1&8,2&7,3&6,4&5,9&0
If x2 + y2 = 3479
Find the number of integral solutions of x,y.
Where x, y < 300.
a) 4
b) 5
c) 0
d) 8
ASHITA MEPANI
BOSTON UNIVERSITY
MOHIT DESAI
SYRACRUSE UNIVERSITY
FATEMA AURANGABADI
NORTHEASTERN UNIVERSITY
BHAVISHA DAWDA
NEW YORK UNIVERSITY
MAITRI BUCH
UNIVERSITY OF TEXAS DALLAS
RAVI HARIANI
TUSCON, ARIZONA
SANKET WALAWALKAR
KELLY SCHOOL OF BUSINESS
AMEY BHAVSAR
NEW YORK UNIVERSITY
KUSHAL THAKKAR
CARNEGIE MELLON UNIVERSITY
ANUP DESHPANDE
NEW JERSEY, NOMURA
ROHIT ZAWAR
INDIANNA UNIVERSITY
SAGAR SUCHAK
UNIVERSITY OF TEXAS DALLAS
DARSHI SHAH
NEW YORK UNIVERSITY
AAMIR LAKDAWALA
KAISERSLAUTERN UNIVERSITY, GERMANY
320 + scoring
students from GRE
King
ARCHANA RAMAKRISHNAN
UNIVERSITY OF CALIFORNIA
Target 320+ in GRE
THANK YOU!

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GREKing: Vedic Maths Concept

  • 1. Target 320+ in GRE 1 GREKing
  • 3.  Number Tree  LCM HCF  Divisibility Rules  Power cycle  Remainder Theorem  Remainder of powers an – bn  Last and Second last digit  Power of Exponents  Euler’s Theorem  Fermet’s Theory  Wilson Theorem  Number Systems (decimal binary)
  • 4. Importance in exams CAT – 4 questions CMAT – 2 questions NMAT – 5 questions
  • 6.
  • 8. 42 = 21 x 2 = 7 x 3 x 2 • Factorize the following numbers: • 42 • 72 • 84 • 65 • 108 • 210
  • 9. • Calculate the LCM and HCF for the following groups of numbers a) 42, 70 b) 18, 24, 60 ,150
  • 10. LCM of nos 60 • Three bells ring after 3, 4 and 5 minutes respectively. If they start ringing together, when • will they ring together again?. a) 50 b) 90 c) 60 d) 42
  • 11. HCF of nos 10 • Three buckets of milk having capacities 40, 30 and 50 L respectively. What is the largest • size of the measure that can be used to pour milk from all three buckets such that the • volumes of milk contained in each bucket is an integral number of pourings used by the • measure. a) 50 b) 10 c) 60 d) 42
  • 12. A = 12 x B = 12 y 144 xy = 2880 xy = 20 1, 20 2, 10 4, 5 5,4 10,2 20,1 • The product of two numbers is 2880. If the HCF is 12, how many such pairs of number are possible? a) 3 b) 6 c) 8 d) 2
  • 13. LCM of nos 720 • Find the least number which when divided by 9,12,16,30 leaves in each case a • remainder of 3. a) 721 b) 722 c) 723 d) 754
  • 14. LCM of nos 15 105 120 135….. • Find the number of all the numbers divisible by 3 and 5 between 100 to 200. a) 9 b) 8 c) 6 d) 7
  • 15. M(3) + M(5) – M(3,5) 30 + 20 – 7 = 43 • Find the number of all the numbers divisible by 3 or 5 between 100 to 200. a) 25 b) 89 c) 60 d) 43
  • 16. LCM of nos 15 105 120 135….. Sn = n/2 (a+l) • Sum of all the numbers divisible by 3 and 5 between 100 to 200. a) 1050 b) 8090 c) 6020 d) 4231
  • 17. 120 • What is the least three digit number that is divisible by 5 and 6? • - What is the least 4 digit number that is divisible by 12 and 15? • - What is the least 4 digit number that is divisible by 25 and 30?
  • 19. TESTS OF DIVISIBILITY: • Divisibility By 2 : A number is divisible by 2, if its unit's digit is any of 0, 2, 4, 6, 8. • Ex. 84932 is divisible by 2, while 65935 is not. • Divisibility By 3 : A number is divisible by 3, if the sum of its digits is divisible by 3. • Ex.592482 is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2) = 30, which is divisible by 3. • But, 864329 is not divisible by 3, since sum of its digits =(8 + 6 + 4 + 3 + 2 + 9) = 32, which is not divisible by 3.
  • 20. Last 3 multiple of 8 • Find the value of K when 425K is divisible by 8.. a) 6 b) 5 c) 0 d) 4
  • 21. Rule of 2 than 9 • A number A4571203B is divisible by 18. Which of the • following values can A and B take ? a) 1, 2 b) 2, 3 c) 6, 8 d) 3, 3
  • 22. Rule of 8 and 11 • Find A and B , if 3765A56682B is divisible by 88?
  • 23. 21 = 2 22 = 4 23 = 8 24 = 16 25 = 32 …… 2 4 8 6 4k Power Cycle
  • 24. 4k = 2, 3, 7, 8 2k = 4, 9 1k = 1, 5, 6, 0 • 1 = 1, 1, 1, … • 2 = 2, 4, 8, 6… • 3 = 3, 9, 7, 1… • 4 = 4, 6, 4, 6 • 5 = 5, 5, 5, 5… • 6 = 6, 6, 6, 6… • 7 = 7, 9, 3, 1… • 8 = 8, 4, 2, 6… • 9 = 9, 1, 9, 1… • 0 = 0, 0, 0, 0
  • 25. • What is the last digit of 360 • What is the last digit of 743 • What is the last digit of 940 • What is the last digit of 56120 • What is the last digit of 26670
  • 26. • Find the unit's digit in • (264)102 + (264)103 a) 0 b) 2 c) 4 d) 20
  • 27. NMAT• Find the unit's digit in • (263)102 x (265)103 a) 5 b) 2 c) 4 d) 20
  • 29. • Rem (X / A) = Rem (Y/A) x Rem (Z/A) • Where y and z are factors of X
  • 30. • Find the remainder of a) 24^3 / 7 b) 72^6 / 12 c) 625^12 / 12
  • 31. • Find the remainder when 2^246 is divided by 7. a) 1 b) 3 c) 7 d) 9 e) none of these
  • 32. 3^2720 • Find the right most non zero digit of (30)2720 • a.1 b.3 c.7 d.9.
  • 34.
  • 35. N + 1 / N is always 1 N – 1 / N is always 1, -1 • Find the remainder when 3560 is divided by 2. a) 5 b) 2 c) 4 d) 1
  • 36. N + 1 / N is always 1 N – 1 / N is always 1, -1 • Find the remainder when 9560 is divided by 2 = • 33560 is divided by 2 = • 5560 is divided by 4 =
  • 37. N – 1 / N is always 1, -1 • Find the remainder when 7560 is divided by 2 = • 31560 is divided by 2 = • 3560 is divided by 4 =
  • 38. 16 / 17 = -1 So 2 – 1 = 1 CAT 2002 • Find the remainder when 2256 is divided by 17 = a) 3 b) 1 c) 7 d) 9.
  • 39. 2.2256 16 / 17 = -1 So 2 – 1 = 1 • Find the remainder when 2257 is divided by 17 = A. 3 B. 2 C. 7 D. 9
  • 41. 541 / 7 = 5 542 / 7 = 25 / 7 = 4 543 / 7 = 20 / 7 = 6 544 / 7 = 2 545 / 7 = 3 546 / 7 = 1 XAT LEVEL • Find the remainder when 54120 is divided by 7 = a) 3 b) 1 c) 7 d) 9.
  • 42. 41 / 6 = 4 42 / 6 = 4 …… CAT 2003 • Find the remainder when 496 is divided by 6 = a) 3 b) 4 c) 7 d) 9
  • 43. •An+Bn • Is always divisible by A + B for all odd n
  • 44. •An - Bn • Is always divisible by A - B for all odd n • Is always divisible by A - B for all even n • Is always divisible by A + B for all even n
  • 45. 351 - 231 = 12 SNAP • Find the remainder when • 3523 - 2323 is divided by 12 a) 3 b) 0 c) 7 d) 9
  • 46. 352 - 232 = 12, 58 • Find the remainder when • 3522 - 2322 is divided by 58 a) 3 b) 0 c) 7 d) 9.
  • 47. 7+6 = 13; 73 – 63 = 127 CAT 2002 • Find the remainder when • 76n – 66n is divided by ? a) 13 b) 127 c) 559 d) All of these
  • 49. 780/7 + 780/49 + 780/343 CMAT • Find the highest power of 7 which will divide 780! a) 13 b) 127 c) 559 d) 15
  • 50. 250/5 + 250/25 + 250/125 • Find the highest power of • 25 which will divide 250! a) 13 b) 127 c) 559 d) 31
  • 51. Wilson's Theorem • Let p be an integer greater than one. p is prime if and only if (p-1)! = -1 (mod p).This beautiful result is of mostly theoretical value because it is relatively difficult to calculate (p-1)! In contrast it is easy to calculate ap-1, so elementary primality tests are built using Fermat's Little Theorem rather than Wilson's. Neither Waring or Wilson could prove the above theorem, but now it can be found in any elementary number theory text. To save you some time we present a proof here. • Proof. It is easy to check the result when p is 2 or 3, so let us assume p > 3. If p is composite, then its positive divisors are among the integers1, 2, 3, 4, ... , p-1 and it is clear that gcd((p-1)!,p) > 1, so we can not have (p-1)! = -1 (mod p). • However if p is prime, then each of the above integers are relatively prime to p. So for each of these integers a there is another such that ab = 1 (mod p). It is important to note that this b is unique modulo p, and that since p is prime, a = b if and only if a is 1 or p-1. Now if we omit 1 and p-1, then the others can be grouped into pairs whose product is one showing2.3.4.....(p-2) = 1 (mod p) (or more simply (p-2)! = 1 (mod p)). Finally, multiply this equality by p-1 to complete the proof.
  • 54. • Let N=1421*1423*1425.What is the remainder when N is divided by 12? a) 0 b) 9 c) 3 d) 6
  • 55. 2^4+2/5=3. • What is the remainder when (2^100 + 2) is divided by 101? a) 3 b) 4 c) 5 d) 6
  • 56. 2^n & 5^n=n+1. • The values of numbers 2^2004 and 5^2004 are written one after another. How many digits are there in all? a) 4008 b) 2003 c) 2005 d) none of these
  • 57. 3^2720=1. • Find the right most non zero digit of (30)^2720. a) 1 b) 3 c) 7 d) 9.
  • 58. A^n + B^n=divisible by a+b • The remainder, when (15^23 + 23^23) is divided by 19 is: a) 4 b) 0 c) 15 d) 18.
  • 59. Divisibility rule of 27 • What is the remainder of 22222222222….(27 times) is divided by 27? a) 0 b) 1 c) 3 d) 4
  • 60. N^n+1 /n=1 • Find the remainder when 2009^2010 is divided by 2011. a) 2010 b) 0 c) 1 d) 2009
  • 61. Understand the format of Q • 111^111 +222^111 +333^111 +444^111+….999^111 is divided by 555? a) 1 b) 0 c) 2 d) 3 e) 4
  • 62. 1! +2*2!/3=rem. is 2 • Let N!=1*2*3*4*5*6*N for N greater or equal to 1. • If P=1! +(2*2!) +(3*3!) +(4*4!)……(12*12!). Find the remainder when P is divided by 13. a) 11 b) 1 c) 2 d) 12
  • 63. • If R=30^65 -29^65/30^64+29^64. then, a) 0<R<1 b) 0.5<R<1 c) 0<0.1 d) R>1
  • 64. 306 is divisible by 3, 6, 9 • 3066 - 306 is not divisible by which of the following ? a) 3 b) 4 c) 6 d) 9
  • 65. We need 1 even and two odd • P,Q& R form a set if distinct prime numbers less than 20.How many such prime numbers are possible for which the sum of P,Q & R is even. a) 15 b) 21 c) 24 d) 28
  • 66. • Q5. (153)8 = (x)10. a) 162 b) 107 c) 166 d) 207 e) 203
  • 67. Find the last two digits if 1733^148.
  • 68. • Finding the last n second last digit • (2^10)even= 24 • (2^10)odd =76
  • 69. • Finding the last n second last digit • For odd digits • (21)^27= 41 • (31)^21=31 • (a1)^cd = ad1
  • 70. Concept of perfect squares examining the nature of all perfect squares… Any number will have a unit’s digit as.. 1,2,3,4,5,6,7,8,9,0. Examining the nature of 1. 1^2=01 11^2=121 21^2=441 31^2=961 n so on… Nature of 2. 2^2=04 12^2=144 22^2=484 32^2=1024 n so on… Nature of 3. 3^2=09 13^2=169 23^2=529 33^2=1089.. Nature of 4 4^2=16 14^2=196 24^2=576 34^2=1156.. Nature of 5 5^2=25 15^2=225 25^2=625 35^2=1225.. Nature of 6 6^2=36 16^2=256 26^2=576.. Nature of 7 7^2=49 17^2=289 27^2=729.. Nature of 8 8^2=64 18^2=324 28^2=784.. Nature of 9 9^2=81 19^2=361 29^2=841..
  • 71. 74= 2401 => last digit 1 78= 5764801 => last digit 1 72008 = 1 , cyclist of four. What is the last two digits of 72008 a) 71 b) 51 c) 01 d) 11
  • 72. X can be even or odd & same for Y. If x2 - y2 = 1234 Find the number of integral solutions of x,y. Where x, y < 150. a) 4 b) 5 c) 0 d) 8
  • 73. To get the last digit as 9,the only ways are 1&8,2&7,3&6,4&5,9&0 If x2 + y2 = 3479 Find the number of integral solutions of x,y. Where x, y < 300. a) 4 b) 5 c) 0 d) 8
  • 74. ASHITA MEPANI BOSTON UNIVERSITY MOHIT DESAI SYRACRUSE UNIVERSITY FATEMA AURANGABADI NORTHEASTERN UNIVERSITY BHAVISHA DAWDA NEW YORK UNIVERSITY MAITRI BUCH UNIVERSITY OF TEXAS DALLAS RAVI HARIANI TUSCON, ARIZONA SANKET WALAWALKAR KELLY SCHOOL OF BUSINESS AMEY BHAVSAR NEW YORK UNIVERSITY KUSHAL THAKKAR CARNEGIE MELLON UNIVERSITY ANUP DESHPANDE NEW JERSEY, NOMURA ROHIT ZAWAR INDIANNA UNIVERSITY SAGAR SUCHAK UNIVERSITY OF TEXAS DALLAS DARSHI SHAH NEW YORK UNIVERSITY AAMIR LAKDAWALA KAISERSLAUTERN UNIVERSITY, GERMANY 320 + scoring students from GRE King ARCHANA RAMAKRISHNAN UNIVERSITY OF CALIFORNIA Target 320+ in GRE