Remainders

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Remainders

by topspin360 » Sat Feb 02, 2013 10:15 am
Can someone please thoroughly explain the logic behind this problem.

17. A number when divided by another number leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?
(A) 13
(B) 59
(C) 35
(D) 37
(E) 12

D

Thanks!
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by Brent@GMATPrepNow » Sat Feb 02, 2013 10:42 am
topspin360 wrote:Can someone please thoroughly explain the logic behind this problem.

17. A number when divided by another number leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?
(A) 13
(B) 59
(C) 35
(D) 37
(E) 12

D
There's a nice rule that say, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2

A number when divided by another number leaves a remainder of 24
We're not told the quotient here (it's implied), so let's add it to our sentence to get:
A number when divided by another number equals k with remainder 24
Let the original number be N and let the divisor be x
So, we get: N divided by x equals k with remainder 24
Using the above rule, we can write: kx + 24 = N

When twice the original number is divided by the same divisor, the remainder is 11.
Once again, the quotient is implied, so let's rewrite this as:
When twice the original number is divided by the same divisor, we get j with remainder 11.
In other words: 2N divided by x equals j with remainder 11
Using our rule, we can write: jx + 11 = 2N

We now have two equations:
kx + 24 = N
jx + 11 = 2N

Multiply the top equation by 2 so that we have 2N in both equations:
2kx + 48 = 2N
jx + 11 = 2N

Since both of the left-hand expressions equal 2N, they must be equal:
2kx + 48 = jx + 11 (our goal is to find the value of x)

Rearrange to get: jx - 2kx = 37
Factor to get: x(j-2k) = 37

IMPORANT: we know that x and (j-2k) are integers. We also know that 37 is prime.
So, if (something)(something) = 37, then one of those somethings must be 1 and the other something must be 37.

So, x equals either 1 or 37.
x cannot equal 1, since N divided by 1 will never leave a remainder.
So, x must equal 37.

Answer = D

Cheers,
Brent
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by chiccufrazer » Sat Feb 02, 2013 11:28 am
Wow nice explaination at brent

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by chiccufrazer » Sat Feb 02, 2013 11:28 am
Wow nice explaination at brent

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by GMATGuruNY » Sat Feb 02, 2013 12:09 pm
topspin360 wrote:Can someone please thoroughly explain the logic behind this problem.

17. A number when divided by another number leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?
(A) 13
(B) 59
(C) 35
(D) 37
(E) 12
When x is divided by y, the remainder is 24.
When 2x is divided by y, the remainder is 11.
We can plug in the answers, which represent the value of y.
When we divide x by y, the remainder of 24 must be LESS than y.
Thus, y must be GREATER than 24.
Eliminate A and E.

Remaining options:
B: y = 59.
C: y = 35.
D: y = 37.

If we add a remainder of 24 to these possible divisors, we get the following possible values for x:
B: x = 59+24 = 83.
C: x = 35+24 = 59.
D: x = 37+24 = 61.

When we divide 2x by y, we get:
B: 2x/y = (2*83)/59 = 166/59 = 2 R48.
C: 2x/y = (2*59)/35 = 118/35= 3 R13.
D: 2x/y = (2*61)/37 = 122/37 = 3 R11.

Only answer choice D yields the required remainder of 11.

The correct answer is D.
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