Division Question - Tough?

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Division Question - Tough?

by mohitv » Sat Mar 30, 2013 12:21 pm
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





Can anybody teach me a simple method to solve this question? I tried solving it using Manhattan method but couldnt crack it. I will post the OA afterwards.

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by Brent@GMATPrepNow » Sat Mar 30, 2013 12:30 pm
mohitv wrote: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
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 rintoo22 » Sat Mar 30, 2013 12:34 pm
If the remainder is 24. Then option A and E are eliminated. That leaves us with B, C, D.
Plug In "divided by another number leaves a remainder of 24" and "twice the original number is divided by the same divisor" Should give the answer.

Answer D.

61/37 gives remainder 24
122/37 gives remainder 11.

However I must admit that there could be a better way to do this.

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by mohitv » Sat Mar 30, 2013 12:57 pm
Hey Brent,
Thanks for the reply.
I also solved it in a similar manner, but I thought the method was not correct.

Anyways, what is be the difficulty level of this question?

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by kul512 » Sat Mar 30, 2013 8:50 pm
Just another way-
if number is doubled, the remainder must also double i.e. remainder will become 48. Now when remainder is 11, then divisor must be 37 to leave 11 after division.
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by GMATGuruNY » Sun Mar 31, 2013 3:38 am
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.
Since 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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by hemant_rajput » Sun Mar 31, 2013 11:58 am
let N be the no. when divided by divisor D leaves remainder 24 and quotient Q.

we can represent N = D*Q + 24.

double the original no.

2N = 2DQ + 48

which is equal to D*Y + 11, here Y is new quotient.

So a no. 2N - 48 or 2N - 11 will be completely divisible by 2DQ or DY.

so now we know that 11 came because when we divide 48 by D we get quotient 1 and remainder 11. so difference between 48 and remainder 11 will be the division which is 37.


Hope this helps.
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.