Numbers #4 : GMAT Problem Solving (PS)

Numbers #4

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Numbers #4 [#permalink]

24 Nov 2011, 10:37

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A number when divided by 36 leaves a reminder of 23. Which of the following could be a reminder when it is divided by72.

A 13 B 59 C 37 D 21

[Reveal] Spoiler: OA

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Re: Numbers #4 [#permalink]

24 Nov 2011, 12:41

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N = 36*a+23. Try a = 1, then N = 59. 59/72 = 0 with remainder 59. Answer is B.
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Re: Numbers #4 [#permalink]

24 Nov 2011, 23:11

Answer is B , but is there a generic formula or approach for this kind of problem instead of trail and error approach ?
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h(n) defined as product of even integers from 2 to n
Number N divided by D leaves remainder R
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Re: Numbers #4 [#permalink]

25 Nov 2011, 04:22

I got B... Used the long approach
List multiples of 36
Add 53 to them and compare result to answers

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Re: Numbers #4 [#permalink]

25 Nov 2011, 11:11

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liftoff wrote:

I got B... Used the long approach
List multiples of 36
Add 53 to them and compare result to answers

liftoff wrote:

I got B... Used the long approach
List multiples of 36
Add 53 to them and compare result to answers

thisiszico2006 wrote:

Answer is B , but is there a generic formula or approach for this kind of problem instead of trail and error approach ?

U need to have an approach to solve this. Lets solve this logiccally

Here is the Approach
Given that N=36k+23 Here k could be either odd or even.
we got to find the reminder when 36 multiplied by a number, is divided by 72. So in the above equation K could be of odd or even numbers. give k values starting from 0 to 4

Case 1. When k=0 (36*0)+23 /72 gives reminder 23
Case 2. when k=1 (36*1)+23 /72 gives reminder 59.How we got59 is that-36/72gives reminder36.This 36+ 23=59. )
Case 3.when k= 2 (36*2)+23 /72 gives reminder 23
Case 4.When k=3 (36*3)+23 /72 gives reminder 59
So from this it can be understood that reminder can be either 23 or 59 based on the value of k if its even or odd.
In the choice only 59 is given.
So the answer is B 59 BINGO!!
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Re: Numbers #4 [#permalink]

25 Nov 2011, 12:57

So the learning from this problem can be summarized below :

If a number "N" is divided by "D" and leaves a remainder "R" , then if that same number "N" is divided by "k*D" where "k" is a positive integer , then the number of remainders that are possible are R , D+R , 2D+R , .... , (k-1)D + R.

So if the question was that what is the remainder when the number was divided by 288 , we can approach it in a mechanical way.

The possible remainders are 23 , 36+23 , 72+23 , .... , 252+23 . We can straightaway reject answer choices that are not in this list.

Sample question : A number N divided by D gives a remainder of 7 , the same number divided by 5D gives a remainder of 24. Find D [N and D are both positive integers].

Ans: The possible remainder for the 2nd case are 7 , D+7 , 2D+7 , 3D+7 , 4D+7.
So 7<> 24, case closed for 1st possibility ,
D+7 = 24 or D=17 (ok)
2D+7= 24, not ok as D cannot be fraction.
3D + 7=24, not ok as D cannot be fraction
4D+ 7 = 24, not possible as D cannot be fraction.

So possible values of D is only 17. So 17 must be present among the answer choices.
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Re: Numbers #4 [#permalink] 25 Nov 2011, 12:57

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Numbers #4

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