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a, b, c, and d are positive integers. If the remainder...

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a, b, c, and d are positive integers. If the remainder...

by BTGmoderatorLU » Sun Dec 17, 2017 11:36 am
a, b, c, and d are positive integers. If the remainder is 9 when a is divided by b, and the remainder is 5 when c is divided by d, which of the following is NOT a possible value for b + d?

(A) 20
(B) 19
(C) 18
(D) 16
(E) 15

The OA is E.

I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
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by DavidG@VeritasPrep » Sun Dec 17, 2017 11:40 am
LUANDATO wrote:a, b, c, and d are positive integers. If the remainder is 9 when a is divided by b, and the remainder is 5 when c is divided by d, which of the following is NOT a possible value for b + d?

(A) 20
(B) 19
(C) 18
(D) 16
(E) 15

The OA is E.

I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.
Important Rule: The divisor is always greater than the remainder. If the remainder is 9 when a is divided by b, then we know that b, the divisor is greater than 9, so the minimum value is 10.

If the remainder is 5 when c is divided by d, we know that d, the divisor, is greater than 5, the minimum value is 6

If the mi value of b is 10 and min value of d is 6, then the min value of b + d is 16. Therefore the sum could never be 15. The answer is E
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by [email protected] » Sun Dec 17, 2017 1:44 pm
Hi LUANDATO,

We're told that A, B, C and D are positive integers, A/B gives us a remainder of 9 and C/D gives us a remainder of 5. We're asked which of the following is NOT a possible value for B+D. This is a great 'concept question' - and you can solve it without doing much math IF you understand the concept that this question is built around.

When dealing with remainders, the possible value of the remainder is 'limited' by what the denominator is. For example:

If you divide an integer by 2, then the ONLY possible remainders will be 0 and 1. (re: 2/2 = 1r0 and 3/2 = 1r1, then the cycle repeats).
If you divide an integer by 3, then the ONLY possible remainders will be 0, 1 and 2. (re: 3/3 = 1r0 and 4/3 = 1r1 and 5/3 = 1r2 then the cycle repeats).
Etc.

The pattern is that the largest possible remainder will be '1 LESS' than the denominator. We can use that concept 'in reverse' to answer this question.

Here, in one calculation we're getting a remainder of 9 (so the denominator must be AT LEAST 10) and in the other calculation we're getting a remainder of 5 (so the denominator must be AT LEAST 6). Thus, the minimum value of B+D is 10+6 = 16. Therefore, 15 is NOT a possible value.

Final Answer: E

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