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Two numbers when divided by a divisor leave reminders of 248 [#permalink]
 23 Nov 2011, 09:14
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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
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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.
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cleetus wrote: 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 Thanks kostyan5. My approach is similar to that of urs. This is how i did it. Let the 2 numbers be X and Y; Let D= Divisor X = D*N+248 , N = Quotient got when X is divided by divisor R Y = D*K+372 , K = Quotient got when Y is divided by divisor R X+Y = (D*N+248) + (D*K+372) = D(N+K)+620 = D(N+K+552/D)+68 As N+K+552/D must be an integer, D must be a factor of 552. As any divisor is greater than the reminder, D>372 So D=552 Answe B
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kostyan5 wrote: 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. ok, "a" must be at least 373, but then why not 414 instead of 552? Thanks!
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Saurajm wrote: kostyan5 wrote: 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. ok, "a" must be at least 373, but then why not 414 instead of 552? Thanks! 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.
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Re: Two numbers when divided by a divisor leave reminders of 248 [#permalink]
31 Jan 2012, 03:24
Once we know that a=552/ (M-N-K),
Can we say that a < or = 552.
And since 372 is one of the remainders (eliminates A. 276) the only possibility is 552 itself.
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kostyan5 wrote: 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. Why did you decide the a must be at least 373 and not 248? That's the other remainder.
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elaskova wrote: kostyan5 wrote: 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. Why did you decide the a must be at least 373 and not 248? That's the other remainder. So, the divisor mus be greater than both remainders, which means that a>372. Also check this: two-numbers-when-divided-by-a-divisor-leave-reminders-of-123645.html#p1036863Hope it helps.
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A rule to solve all similar problems ---
If two numbers, say a & b, are divided by the same divisor (d) leaving remainders r1 & r2.
Then the remainder (R), when Sum (a+b) / d = (r1+r2) - d.
Note - If R becomes negative, then R = (r1+r2) only.
Hence Solution to the above problem -
d = 68, r1 = 248, r2 = 372
so Remainder R when Sum (a+b) / 68 = (248+372) - 68 = 620 - 68 = 552
Note - Difference (a-b) is exactly divisible by the same divisor (d).
Hope it helps.
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