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Remainder and Divisor

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by harsh.champ » Tue Feb 09, 2010 5:02 am
Two different numbers when divided by the same divisor left remainders of 11 and 21 respectively. When the numbers' sum was divided by the same divisor, the remainder was 4. What was the divisor?

(A)36
(B)28
(C)12
(D)9
(E)None of these

The OA is B
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by shashank.ism » Tue Feb 09, 2010 5:08 am
harsh.champ wrote:Two different numbers when divided by the same divisor left remainders of 11 and 21 respectively. When the numbers' sum was divided by the same divisor, the remainder was 4. What was the divisor?

(A)36
(B)28
(C)12
(D)9
(E)None of these

The OA is B
Let the divisor be a.
x = a*n + 11 ---- (1)
y = a*m + 21 ----- (2)
also given, (x+y) = a*p + 4 ------ (3)
adding the first 2 equations. (x+y) = a*(n+m) + 32 ----- (4)
equate (3) and (4).
a*p + 4 = a*(n+m) + 32
or
a*p + 4 = [a*(n+m) + 28] + 4
cancel 4 on both sides.
u will end up with.
a*p = a*(n+m) + 28.
which implies that 28 should be divisible by a.

Ans b
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by ajith » Tue Feb 09, 2010 6:46 am
harsh.champ wrote:Two different numbers when divided by the same divisor left remainders of 11 and 21 respectively. When the numbers' sum was divided by the same divisor, the remainder was 4. What was the divisor?

(A)36
(B)28
(C)12
(D)9
(E)None of these

The OA is B
n = 11+21 -4 = 28
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