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Integers

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by diehard_gmat » Thu Feb 03, 2011 7:55 am
Letters A, B, C, and D represent four different digits selected from 1, 2, ..., 9. If (A+B)/(C +D) is an integer that is as large as possible, what is the value of A + B?

A) 3
B) 5
C) 11
D) 15
E) 17
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by Anurag@Gurome » Thu Feb 03, 2011 7:58 am
diehard_gmat wrote:Letters A, B, C, and D represent four different digits selected from 1, 2, ..., 9. If (A+B)/(C +D) is an integer that is as large as possible, what is the value of A + B?

A) 3
B) 5
C) 11
D) 15
E) 17
For (A + B)/(C + D) to be a maximum possible integer, the value of (A + B) must be as large as possible and (C + D) must be the smallest positive factor of (A + B) which can be expressed as (C + D).

Now maximum value of (A + B) = (8 + 9) = 17
But smallest factor of 17 is 1, which can't be expressed as (C + D)

Next maximum value of (A + B) = (7 + 8) = 15
Smallest possible value of 15 is 3 = (1 + 2)

Hence, this set of values makes the expression largest integer.

The correct answer is D.
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by monge1980 » Tue Feb 15, 2011 4:14 am
Hi guys can you help me to understand the attached question? Why the answer is A)...
Thanks,
ES
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by Night reader » Tue Feb 15, 2011 4:36 am
(A+B)=i*(C+D) where i is integer as much as possible; (A+B) the largest possible is (9+8)=17 prime. Next large one to these numbers' sum is (9+7)=16; 16=i*(C+D) ---> i is maximized when (A+B) is minimized. So (A+B) can be (1+3)=4 and 16/4=4 OR i=4. Let's try another pair of numbers, as we can also get i=5, when (A+B)=(1+2)=3 and (C+D)=15 or (C+D)=(8+7)=(9+6). Pick D

diehard_gmat wrote:Letters A, B, C, and D represent four different digits selected from 1, 2, ..., 9. If (A+B)/(C +D) is an integer that is as large as possible, what is the value of A + B?

A) 3
B) 5
C) 11
D) 15
E) 17
Last edited by Night reader on Tue Feb 15, 2011 5:02 am, edited 1 time in total.
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by Night reader » Tue Feb 15, 2011 4:53 am
@monge1980: it says 450y=n^3 correct? Also n,y>0
So prime factorization of 450=2^1*3^2*5^2 and we know that 450y is n^3 where n is +ve integer. The only way 450y can be equal to n^3 is when the missing parts in 450 itself can be represented as (...)^3 OR put this 450*(2^2*3^1*5^1). So we must have at least y/(2^2*3^1*5^1) condition, where this ratio is both +ve and integer (at least). Otherwise 450y can't be equal to n^3 or a^3, actually can't be derived the root power of 3

2^2*3^1*5^1 is the part of y here, which completes 410 and help our statement 410y=n^3, because Root^3(410) isn't good, and we look for possible Root^3(410y)...

ok? :)
monge1980 wrote:Hi guys can you help me to understand the attached question? Why the answer is A)...
Thanks,
ES
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by monge1980 » Tue Feb 15, 2011 5:28 am
Thanks!!
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