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Marge has n candies

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Marge has n candies

by sanju09 » Tue Nov 15, 2011 1:34 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 shankar.ashwin » Tue Nov 15, 2011 1:40 am
Let N be the total candies.

N = 5a + 2
N = 6b + 1

6b - 5a = 1 ( 5a can be 25,30,35,40,45) Find a multiple of 6 which is 1 more than multiple of 5,

Only 36,35 satisfy this condition.

So total = 37 candies, 37/7 -> remainder = 2 C IMO
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by neelgandham » Tue Nov 15, 2011 2:25 am
sanju09 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.
n can be one among 22,27,32,37,42,47
If she divides the candies among 6 children, she will have 1 candy remaining.
n is 37
How many candies will remain if she divides the candies among 7 children?
37/7, remainder = 2, IMO C
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