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the candies among 7 children

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by sanju09 » Tue Sep 14, 2010 4:08 am
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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by Rahul@gurome » Tue Sep 14, 2010 5:23 am
When Marge divides n candies among 5 children let each child gets x candies.
Then, n = 5x + 2
When Marge divides n candies among 6 children let each child gets y candies.
Then, n = 6y + 1
So, we have 5x + 2 = 6y + 1 or 6y - 5x = 1. Now choose the values of x and y so that their difference is 1, and 20 < n< 50.
Possible integer values are:
x = 7, y = 6 implies 6(6) - 5(7) = 1
n = 5(7) + 2 = 6(6) + 1 = 37

If Marge divides 37 candies among 7 children, then 2 candies will be left.

The correct answer is [spoiler](C)[/spoiler].
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by ankur.agrawal » Tue Sep 14, 2010 5:28 am
Alternate way particularly for this question:

equally among 5 children = we can have 22,27,32,37,42,47 as possible values.

equally among 6 children = we can have 25,31,37,42,47 as possible values.

So only 37 is possible . So n =37.

Dividing by 7 we get 2 as remainder.

C
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