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Marge has n candies, where n is an integer such that 20 < n< 50. If Ma

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Marge has n candies, where n is an integer such that 20 < n< 50. If Ma [#permalink]

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Marge has n candies, where n is an integer such that 20 < n< 50. If Marge divides the candies equally among 5 children, she will have 2 candies remaining. If she divides the candies among 6 children, she will have 1 candy remaining. How many candies will remain if she divides the candies among 7 children?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

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Re: Marge has n candies, where n is an integer such that 20 < n< 50. If Ma [#permalink]

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Bunuel wrote:
Marge has n candies, where n is an integer such that 20 < n< 50. If Marge divides the candies equally among 5 children, she will have 2 candies remaining. If she divides the candies among 6 children, she will have 1 candy remaining. How many candies will remain if she divides the candies among 7 children?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4



If Marge divides the candies equally among 5 children, she will have 2 candies remaining.
This tells us that the candies are of 5x + 2 type and so 22, 27, 32, 37, 42, or 47

If she divides the candies among 6 children, she will have 1 candy remaining.
this tells us it is if 6x+1 type and so 25, 31, 37, 43, or 49

ONLY 37 is common, so candies are 37 in number..
if 37 is divided in 7 kids, remainder = 2

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Re: Marge has n candies, where n is an integer such that 20 < n< 50. If Ma [#permalink]

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New post 11 Oct 2017, 08:30
I got it wrong because i selected wrong option in hurry.
Here goes the sol.

Let the total candies be x
Now when x is divided by 5, remainder is 2 => either the unit digit of x is 2 or 7 => x can be 22, 27, 32, 37, 42, 47.
second condition says, x divided by 6, the remainder is 1.
Only x=37 fulfills the second condition.

Hence, when x is divided by 7, remainder will be 2.
Re: Marge has n candies, where n is an integer such that 20 < n< 50. If Ma   [#permalink] 11 Oct 2017, 08:30
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Marge has n candies, where n is an integer such that 20 < n< 50. If Ma

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