selim wrote:

when a positive integer X is divided by 5 the remainder is 1. when X is divided by 8 the remainder is 4 . what is the smallest positive integer y such that, (X+Y) is divisible by 40 ?

a)3

b)4

c)9

d)13

e) none of these

looking for short-cut solution.

Please post question with correct topic name as first few words of the Question and with OA

x+y has to be divisible by 40, then x+y has to be EVEN..

x is even as it leaves a Remainder of 4 when div by 8..

So y has to be even.. only 4 is possible other than none of these

Eliminate a,c,d

For proper orLogical method..When divided by 5, it leaves a Remainder of 1.

So x will have units digit 1 or 6

When divided by 8, x leaves a remainder of 4..

So units digit has to be even and thus 6..

Look for 6-4=2 as units digit in MULTIPLE of 8..

First such is 8*4=32..

So x is 32+4=36

Next number after 36 divisible by 40 is 40 itself..

So x+y=40....36+y=40...y=40-36=4

B

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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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