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Hard Remainder Prob. #68 p.70 Quant 2nd Edition

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Hard Remainder Prob. #68 p.70 Quant 2nd Edition

by OneTwoThreeFour » Wed Jan 26, 2011 7:42 pm
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 find the solution in the back overtly confusing. I was able to get to 7n= Q1 + 7/35 and 5n/35 = Q2 + 15/35. Then I combined them to get 12n/35= Q1 + Q2 + 22/35. Not sure if this is the right framework, or I am missing something to keep going on.

Thanks
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by Everest » Wed Jan 26, 2011 8:14 pm
dividend = divisor * quotient + remainder

Let us say N is dividend

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


least value of K for which K+N/35 will be multiple of 35 is 4 since (4+31)/35 =1

4 is the least value of K.
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by OneTwoThreeFour » Wed Jan 26, 2011 8:25 pm
Thanks! There is a pretty good solution in the thread below this one. Picking numbers is still the fastest method though.
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by Night reader » Thu Jan 27, 2011 5:01 am
Everest wrote:dividend = divisor * quotient + remainder

Let us say N is dividend

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


least value of K for which K+N/35 will be multiple of 35 is 4 since (4+31)/35 =1

4 is the least value of K.
I guess we have to write here 5x+1 and 7y+3 --> 5x+1=7y+3 --> 5x-2=7y plug integers for y
> x and y positive integers
y=1 --> 5x-2=7
y=2 --> 5x-2=14
y=3 --> 5x-2=21
y=4 --> 5x-2=28 --> x=6

5x+1=7y+3 OR 5*6+1=7*4+3, n=31

k+n=35 for (k+n)/35 --> k+31=35, k=4
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