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When positive integer n is divided by 5, the remainder is 1.

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Re: When positive integer n is divided by 5, the remainder is 1.  [#permalink]

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New post 22 Feb 2018, 13:55
Bunuel wrote:
SOLUTION

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

Positive integer n is divided by 5, the remainder is 1 --> \(n=5q+1\), where \(q\) is the quotient --> 1, 6, 11, 16, 21, 26, 31, ...
Positive integer n is divided by 7, the remainder is 3 --> \(n=7p+3\), where \(p\) is the quotient --> 3, 10, 17, 24, 31, ....

There is a way to derive general formula for \(n\) (of a type \(n=mx+r\), where \(x\) is divisor and \(r\) is a remainder) based on above two statements:

Divisor \(x\) would be the least common multiple of above two divisors 5 and 7, hence \(x=35\).

Remainder \(r\) would be the first common integer in above two patterns, hence \(r=31\).

Therefore general formula based on both statements is \(n=35m+31\). Thus the smallest positive integer k such that k+n is a multiple of 35 is 4 --> \(n+4=35k+31+4=35(k+1)\).

Answer: B.

More about deriving general formula for such problems at: http://gmatclub.com/forum/manhattan-rem ... ml#p721341



what is the logic behind these numbers :? -> 1, 6, 11, 16, 21, 26, 31, ...
Re: When positive integer n is divided by 5, the remainder is 1. &nbs [#permalink] 22 Feb 2018, 13:55

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