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OG GMAT Quantative Review - PS #68

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by arzanr » Sat Feb 27, 2010 7:44 pm
poweroc wrote: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?

A - 3
B - 4
C -12
D - 32
E - 35

I have a lot of trouble with these types of questions and the OG answer is quite confusing. Thanks
k+n = 35, therefore, n = 35 - k

Any multiple of 35 would be divisible by 5 therefore k cannot be 3 because n needs to be a multiple of 5 - 1. E.g. if you take the smallest multiple of 35, i.e. 35 itself you can see that if n + k =35 and since we know n-1 is a multiple of 5, then k can't be 3 because that would make n-2 a multiple of 5. If you try with 4 you see that n would be equal to 31 and it fits both criteria, n-1 is a multiple of 5 (30) and n-3 is a multiple of 7 (28).
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Source: — Problem Solving |

by devp » Sun Feb 28, 2010 8:43 am
The number n is such that it has to satisfy these two series
a. 5m + 1
b. 7j + 3

To solve these types first take the LCM of the multiples i.e 5 and 7 here = 35.
So the resulting series will be
35i + r.

Now we find r. To do this set i=0 and try to find the first term in the series a and b that is same.
series b. 10,17,24,31,38
series a. 6,11,16,,,31 So it matches at 31. So r=31

Therefore the resulting series is :
35i + 31. So the smallest number which needs to be added to this is 4 to make it divisible by 35.

Therefore B.
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by tcon » Fri Oct 14, 2011 6:47 am
Im still confused by this explaination - can a gmat instructor break it down simply please?
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by GMATGuruNY » Fri Oct 14, 2011 6:55 am
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by HeyArnold » Sun Oct 16, 2011 7:50 am
n / 5 has remainder of 1

So what divided by 5 has remainder of 1?
6, 11, 16, 21, 26, 31...

n / 7 has remainder of 3

So what divided by 7 has a remainder of 3?
10, 17, 24, 31...

31 satisfies both elements.

So the smallest k where K + n is a multiple of 35 is 4, where 31 + 4 = 35, the smallest multiple of 35.

B
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