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when a positive integer X is divided by 5 the remainder is 1.

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when a positive integer X is divided by 5 the remainder is 1. [#permalink]

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New post 23 Jan 2018, 15:10
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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.
[Reveal] Spoiler: OA

Last edited by chetan2u on 23 Jan 2018, 17:39, edited 3 times in total.
updated topic name and added OA
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Re: when a positive integer X is divided by 5 the remainder is 1. [#permalink]

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New post 23 Jan 2018, 15:35
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The first thing I did when attempting this problem is to find the lowest value that satisfies the information given.
We can quickly compare some values (6, 11, 16, 21, 26, 31, 36, 41) and (4, 12, 20, 28, 36, 44) and find out quickly that 36 is the lowest value which fits both the facts we are given.

from 36 we can look at the answer choices and see that 4 can be added to make it 40.
Answer is b)
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Re: when a positive integer X is divided by 5 the remainder is 1. [#permalink]

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New post 23 Jan 2018, 17:35
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

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Re: when a positive integer X is divided by 5 the remainder is 1. [#permalink]

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New post 23 Jan 2018, 19:15
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.


x=5q+1
x=8p+4
8p+3=5q
least value of p that makes q an integer is 4
p=4; q=7; x=36
40-36=4
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Re: when a positive integer X is divided by 5 the remainder is 1. [#permalink]

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New post 23 Jan 2018, 19:41
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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.


This is a copy of the following OG question: https://gmatclub.com/forum/when-positiv ... 66831.html
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Re: when a positive integer X is divided by 5 the remainder is 1.   [#permalink] 23 Jan 2018, 19:41
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