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jain4444 commented on a post

1h

Let N be the number that consists of 61 consecutive 3's, so N=333…..61 times. Let M be the number that consists of 62 consecutive 6's, so M=6666…..62 times. What is the digital sum of N×M?

20 Comments

Angadbir Digit Sum = 9 * 62 = 558 Digital Sum = 9 . 1h.

jdsnk59 @Ibanez @Angadbir Didn't understand how you found out 558?. 1h.

jain4444 replied to Official Quant thread for CAT 2013

jain4444

1h

Let N be the number that consists of 61 consecutive 3's, so N=333…..61 times. Let M be the number that consists of 62 consecutive 6's, so M=6666…..62 times. What is the digital sum of N×M?

20 Comments

Angadbir Digit Sum = 9 * 62 = 558 Digital Sum = 9 . 1h.

jdsnk59 @Ibanez @Angadbir Didn't understand how you found out 558?. 1h.

jain4444 replied to Official Quant thread for CAT 2013

jain4444

3h

Find the number of ordered pairs of distinct positive primes p, q (p≠q) such that p^2 + 7pq + q^2 is the square of an integer.

7 Comments

JanardanJakad My approach (p-q)^2 + 9pq = N^2 =>(N+p-q)(N-p+q) = 9p.... 4h.

amaze11 Total cases : 12 (p+q)^2+5pq=m^2 (6 cases) & (p-q)^2+.... 3h.

jain4444

05:32 PM, 22 May '13

Official Quant thread ...

3h

Find the number of ordered pairs of distinct positive primes p, q (p≠q) such that p^2 + 7pq + q^2 is the square of an integer.

RC practice :- http://codecoax.com/grerc/

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JanardanJakad options
#1 • 5h ago Like
jayasrivastava0 2 pairs..(3,11) &(11,3)
#2 • 4h ago Like
jayasrivastava0
let p^2 +q^2 +7pq=m^2
5pq=m^2-(p+q)^2= (m +p+q)(m−p−q)
Case 1
m+p+q = 5p and m−p−q = q
=>2(p+q) = 5p−q. leads to p=q
case 2
m + p+ q = 5q and m − p − q = p,leads to p = q
case 3
m+p+q = pq, m−p−q = 5
=>2(p+q) = pq−5
=>(p − 2)(q − 2) = 9
(5,5),(3,11),(11,3) satify the above condition.

(p≠q), so only two are possible.

#3 • 4h ago Like 1
JanardanJakad 2 pairs (11,3) and (3,11)
#4 • 4h ago Like
jaffa (P+q)^2+5pq = K^2
K^2-(P+q)^2=5pq
(3,11) and (11,3) satisfies ...
#5 • 4h ago Like
JanardanJakad My approach

(p-q)^2 + 9pq = N^2
=>(N+p-q)(N-p+q) = 9pq

4 cases:

Case 1:
N+p-q = 9p
N-p+q = q

Case 2:
N+p-q = 9q
N-p+q = p
Case 3:
N+p-q = 9
N-p+q = pq
Case 4:
N+p-q = pq
N-p+q = 9

Taking Case 1:
Subtracting the two, we get 2p - 2q = 9p - q => Invalid as ratio comes in negative

Taking Case 2:
Subtracting we get 2p - 2q = 9q - p => p/q = 11/3

Similarly we will do for Case 3 and Case4 where we will find that the solution exists when p =q = 3

Hence only 2 pairs exist (11,3) and (3,11)

#6 • 4h ago Like
amaze11 Total cases : 12
(p+q)^2+5pq=m^2 (6 cases) & (p-q)^2+9pq=m^2

Considering (p+q)^2+5pq=m^2 => (m-p-q)(m+p+q)=5pq
CASE 1 : m-p-q = 5 & m+p+q=pq
CASE 2 : m-p-q = 5p & m+p+q=q
CASE 3 : m-p-q = 5q & m+p+q=p
CASE 4 : m-p-q = p & m+p+q=5q
CASE 5 : m-p-q = q & m+p+q=5p
CASE 6 : m-p-q = pq & m+p+q=5

And similarly 6 cases for other equation.
Answer, however, is still (3,11), (11,3)

#7 • 3h ago Like

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jain4444 commented on a post

jain4444

6h

Equilateral triangle ABC has a circumcircle X with center O and circumradius 10. Another circle X1 is drawn inside X such that it is tangential to radii OC and OB and circle X. The radius of X1 can be expressed in the form a*rootb − c , where a,b and c are positive integers, and b is not divisibl...

bmp file

12 Comments

jain4444 sir O1Y toh r hai (25 - r) OO1 hai but wo 5 + (5 - r) = .... 8h.

jain4444 answer is 53 . 8h.

jain4444 commented on a post

anupam7

10h

How many sets of three numbers (triplets) possible for LCM = 72 ??

5 Comments

jain4444 72 = 2^3 * 3^2 x = 2^x1 * 3^x2 y = 2^y1 * 3^y2 z = 2.... 10h.

jain4444 replied to Official Quant thread for CAT 2013

jain4444

6h

Equilateral triangle ABC has a circumcircle X with center O and circumradius 10. Another circle X1 is drawn inside X such that it is tangential to radii OC and OB and circle X. The radius of X1 can be expressed in the form a*rootb − c , where a,b and c are positive integers, and b is not divisibl...

bmp file

12 Comments

jain4444 sir O1Y toh r hai (25 - r) OO1 hai but wo 5 + (5 - r) = .... 8h.

jain4444 answer is 53 . 8h.

jain4444 commented on a post

abhietwup

2d

The no of ways of factorizing 91000 into two factors m and n,such that m>1, n>1 and gcd(m,n)=1 is

1) 7
2) 15
3) 32
4) none of these

17 users have answered this question.

1 11 Comments

jain4444 commented on a post

enfer

6d

If the sum of two numbers is 480 and their HCF is 12, then How many such pairs exist ?

1 30 Comments

jain4444 12x + 12y = 480 => x + y = 40 now if x is co-prime .... 6d.

jain4444 commented on a post

jain4444

6d

A number is chosen at random from the set {1,2,3,....1998}. K is the probability that it has no prime factors in common with 10! ; then [1/k] = ? [.] = G.I.F

6 Comments

jain4444 10! sir . 6d.

jain4444 replied to Official Quant thread for CAT 2013

jain4444

6d

A number is chosen at random from the set {1,2,3,....1998}. K is the probability that it has no prime factors in common with 10! ; then [1/k] = ? [.] = G.I.F

6 Comments

jain4444 10! sir . 6d.

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