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# When a positive integer x is divided by 5

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Intern
Joined: 13 Aug 2018
Posts: 10
When a positive integer x is divided by 5  [#permalink]

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16 Oct 2018, 06:43
1
3
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Difficulty:

15% (low)

Question Stats:

84% (02:24) correct 16% (02:34) wrong based on 52 sessions

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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 the above

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Joined: 19 Sep 2018
Posts: 1
Re: When a positive integer x is divided by 5  [#permalink]

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16 Oct 2018, 06:56
When it is divided by 5 it leaves remainder 1 so the value of x can be 6,11,16,21,26,31,36,41
and when it is divided by 8 it leaves remainder 4 so he value of x can be
12,20,28,36
so the common value is 36 so from option when we will add 4 it is completely divisible by 40

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Re: When a positive integer x is divided by 5  [#permalink]

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17 Oct 2018, 18:05
jackfr2 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 the above

Posted from my mobile device

In the first instance, we see that x can be values such as 1, 6, 11, 16, 21, 26, 31, 36, etc.

In the second instance, we see that x can be values such as 4, 12, 20, 28, 36, etc.

We see that the first common value is 36, so the smallest value of y is 4 since 36 + 4 = 40.

Alternate Solution:

We have x = 5k + 1 for some integer k and x = 8s + 4 for some integer s. We notice that x + 4 = 5k + 5 = 8s + 8; thus x + 4 is divisible by both 5 and 8. The smallest possible value of x + 4 is 40 since the LCM of 5 and 8 is 40. Then x = 36 and the smallest value of y such that x + y is divisible by 40 is 4.

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Re: When a positive integer x is divided by 5  [#permalink]

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20 Oct 2018, 08:12
jackfr2 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 the above

Posted from my mobile device

OA : B

When a positive integer x is divided by 5 the remainder is 1

x: {1,6,11,16,21,26,31,36,41,46,51,56,61,66,71,76,81...........}

and

when x is divided by 8 the remainder is 4

x: {4,12,20,28,36,44,52,60,68,76,84................}

Given than x+y=40, As we have to find the minimum value of y, x has to 36.

36+y=40

y=4
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Re: When a positive integer x is divided by 5   [#permalink] 20 Oct 2018, 08:12
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