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16 Oct 2018, 07:43
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
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Question Stats:
78% (02:08) correct
22%
(02:24)
wrong
based on 134
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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
Posted from my mobile device
(A) 3
(B) 4
(C) 9
(D) 13
(E) none of the above
Posted from my mobile device
16 Oct 2018, 07:56
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Answer is 4
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
So answer is 4
Posted from my mobile device
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
So answer is 4
Posted from my mobile device
17 Oct 2018, 19:05
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jackfr2
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.
Answer: B
