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Two numbers when divided by a divisor leave reminders of 248 [#permalink]
23 Nov 2011, 08:14

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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.

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.

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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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.

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!

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.

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.

Positive integer a divided by positive integer d yields a reminder of r can always be expressed as a=qd+r, where q is called a quotient and r is called a remainder, note here that 0\leq{r}<d(remainder is non-negative integer and always less than divisor).

So, the divisor mus be greater than both remainders, which means that a>372.

Re: Two numbers when divided by a divisor leave reminders of 248 [#permalink]
14 Nov 2013, 01:30

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Re: Two numbers when divided by a divisor leave reminders of 248 [#permalink]
17 Dec 2013, 14:33

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

Easy, why 68 if the sum of the remainders is 248+372=620? Cause the divisor is eating the other part. Then the divisor is 620-68=552

Answer is A Cheers! J

gmatclubot

Re: Two numbers when divided by a divisor leave reminders of 248
[#permalink]
17 Dec 2013, 14:33