# Candy and boxes

We have some number of candy and empty boxes. When we put candies in boxes of ten, there will be 2 candies and 8 empty boxes left, when of eight, there will be 6 candies and 3 boxes left. How many candy and empty boxes left when we put candies in boxes of nine?

Result

x =  51
y =  11

#### Solution:

$\ \\ b - (k-8) \cdot \ 10=2 \ \\ b - (k-3) \cdot \ 8=6 \ \\ \ \\ b-10k=-78 \ \\ b-8k=-18 \ \\ \ \\ b=222 \ \\ k=30 \ \\ \ \\ b - (k-y) \cdot \ 9=x \ \\ 222 - (30-y) \cdot \ 9=x \ \\ x=9n+6 \ \\ y=n+6 \ \\ n >=2 \ \\ 9n+6 <=222 \ \\ n<=24 \ \\ 2<=n <=24 \ \\ n=5 \ \\ x=9 \cdot \ n+6=9 \cdot \ 5+6=51$
$y=n+6=5+6=11$

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this didnt  help me

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