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23 Nov 2011, 09:14
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Difficulty:
45%
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Question Stats:
72% (02:20) correct
28%
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Two numbers when divided by a divisor leave reminders of 248 and 372 respectively. The reminder obtained when the sum of the numbers is divided by the same divisor is 68. Find the divisor.
A. 276
B. 552
C. 414
D. 1104
E. 2202
A. 276
B. 552
C. 414
D. 1104
E. 2202
23 Nov 2011, 09:57
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Fun question.
Say the two numbers are x and y, and divisor is a.
x divided by a leaves a remainder of 248. This means that x = a*N + 248, where N is the integer result of the division.
y divided by a leaves a remainder of 372. This means that x = a*K + 372, where K is the integer result of the division.
x+y divided by a leaves a remainder of 68. This means that x = a*M + 68, where M is the integer result of the division.
From definitions above:
x+y = (a*N + 248) + (a*K + 372) = a*(N+K) + 620.
a*(N+K) + 620 = a*M + 68
552 = a*(M-N-K)
We know that M, N, and K are integers and that a must be at least 373 (to leave a 372 remainder). The only possible value for (M-N-K) is 1.
Therefore, a = 552. B.
Say the two numbers are x and y, and divisor is a.
x divided by a leaves a remainder of 248. This means that x = a*N + 248, where N is the integer result of the division.
y divided by a leaves a remainder of 372. This means that x = a*K + 372, where K is the integer result of the division.
x+y divided by a leaves a remainder of 68. This means that x = a*M + 68, where M is the integer result of the division.
From definitions above:
x+y = (a*N + 248) + (a*K + 372) = a*(N+K) + 620.
a*(N+K) + 620 = a*M + 68
552 = a*(M-N-K)
We know that M, N, and K are integers and that a must be at least 373 (to leave a 372 remainder). The only possible value for (M-N-K) is 1.
Therefore, a = 552. B.
31 Jan 2012, 02:21
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Saurajm
If we follow kostyan5's way we get 552=a*(M-N-K) --> (M-N-K)=integer=552/a, no other value from the answer choices will yield an integer for this expression except 552 and 276, and as a>372 then a=552.
Hope it's clear.
