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A number when divided by a divisor leaves a remainder of 24. [#permalink]
07 Apr 2007, 22:45

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Difficulty:

25% (low)

Question Stats:

74% (02:23) correct
25% (01:36) wrong based on 116 sessions

A number when divided by a divisor 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) 12 B) 13 C) 35 D) 37 E) 59

I'm already familiar with the textbook method. I'm trying to discover what's wrong with the below method.

I am sorry, but described steps have mistakes in it.
My quick way of solving this would be:

1) a= d+24
2) 2a=kd+11
3) multiply step 1 by 2 and subtruct 1 from 2
4) 0=d(k-2)-37
5) because 37 is prime number, eq would make sense if k=3, hence answer would be 37 (D)

I am sorry, but described steps have mistakes in it. My quick way of solving this would be:

1) a= d+24 2) 2a=kd+11 3) multiply step 1 by 2 and subtruct 1 from 2 4) 0=d(k-2)-37 5) because 37 is prime number, eq would make sense if k=3, hence answer would be 37 (D)

Hi, Thanks. I know. Hence the answer. Will you let me know what mistakes you found.
Also, how does a = d+24 when the prob says "a number when divided by a divisor leaves a remainder of 24"? Is this some sort of shortcut?

A number when divided by a divisor 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

I'm already familiar with the textbook method. I'm trying to discover what's wrong with the below method.

A number when divided by a divisor 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. 6 B. 7 C. 5 D. 8 E. 18

We know that: . The excess remainder is a multiple of the divisor. . The divisor is greater than the greatest remainder (which is 24).

The only multiple of 37 greater than 24 is 37. Therefore, the divisor is 37.

Lets illustrate this by picking numbers: a=24 remainder equation: a/d=k+r/d -a. 24/37=0+24/37 ->remainder is 24 -ax2. 48/37=1+11/37 ->remainder is 11

Another way to see it is through algebra: -a. a/d=k+24/d -> a=dk+24 -ax2. 2a/d=q+11/d -> a=(dq+11)/2 -> 2dk+48=dq+11 Keep in mind that dq = 2dk + "excess remainder" -> 2dk + 48 = 2dk + "excess remainder" + 11 -> "excess remainder" = 37

Sorry if this is not clear enough. Can't think of another way to explain it.

You may want to refer to the Man NP guide for looking into arithmetic with remainders (pg128).

n = da + 24 (1) and 2n = db +11 (2) subtract (1) form (2) you get n = d(b-a) - 13 (3) subtract (1) from (3), you get 0 = d(b-a-a) - 37 it is a prime so lowest d is 37

divisibility and remainders [#permalink]
12 Jul 2011, 08:42

A number when divided by a divisor 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.

I solve till a = fx + 24 2a = fx + 11

After this step I'm lost.

Can someone please explain me the solution, I read the solution but was not totally convinced.

Re: divisibility and remainders [#permalink]
12 Jul 2011, 08:48

mustu wrote:

A number when divided by a divisor 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.

I solve till a = fx + 24 2a = fx + 11

After this step I'm lost.

Can someone please explain me the solution, I read the solution but was not totally convinced.

Regards, Mustu

Looking at the options i know A and E are out ...

so it has to be B ,C or D

59/35 = 24 R 118/35 = 13 R...............wrong

37+24 = 61 61/37 = 24 R 122/37 = 11 R............. right

Re: divisibility and remainders [#permalink]
13 Jul 2011, 12:58

I used N=DQ+24, 2N=DQ+11. Multiplied the first equation by 2 to get 2N on both sides, then set the equations equal to each other. 2DQ+48=DQ+11. Ended up with DQ=-37. Picked D, got it right. Can anyone explain what was wrong with that approach?

gmatclubot

Re: divisibility and remainders
[#permalink]
13 Jul 2011, 12:58