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Remainder Question

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cssears
 Post subject: Remainder Question
 Post Posted: Wed May 12, 2010 10:28 am 
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Course Students


Posts: 15
What is the easiest way to solve a problem such as

when positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer K such that K + N is a multiple of 35?

3
4
12
35
40
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jinkala2
 Post subject: Re: Remainder Question
 Post Posted: Mon May 17, 2010 1:28 pm 
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Forum Guests


Posts: 2
N/5 = 1 ; N/7 = 3

so 5x+1 = N 7x+3 =N ; x=1,2,3.......

substitute x value such that
5x+1 = N -- 5*6+1 = 31
7x+3 = N -- 7*4+3 = 31

(K+N) / 35 must be equal to 1 to find the least value of k

if k=4 --- (4+31)/35 =1

hence k=4
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mschwrtz
 Post subject: Re: Remainder Question
 Post Posted: Thu Oct 28, 2010 5:36 pm 
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ManhattanGMAT Staff


Posts: 506
That is probably the best way for many test-takers, but others may prefer to consider particular values for n.

when positive integer n is divided by 5, the remainder is 1. In other words, n is one more than some multiple of 5:

1
6
11
16
21
.
.
.
anything ending with a 1 or a 6.

When n is divided by 7, the remainder is 3. In other words, n is 3 more than some multiple of 7:
3
10
17
24
31
.
.
.
31 is the first number in that list that ends with a 1 or a 6. So n could be 31

What is the smallest positive integer K such that K + N is a multiple of 35? For n=31, the smallest such integer is 4. that eliminates every answer but A and B. If you're concerned that some other value for K might give a different result, just keep adding 7s to the list above until you see another K that ends with a 1 or a 6. The next such K will be 66, which also yields 4.

It's not quite as neat as the algebraic solution, but it's relatively easy to implement.
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