# 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?

Correct result:

x =  51
y =  11

#### Solution:

$\ \\ b - (k-8) \cdot \ 10=2 \ \\ b - (k-3) \cdot \ 8=6 \ \\ \ \\ b-10k=-78 \ \\ b-8k=-18 \ \\ \ \\ b=222 \ \\ k=30 \ \\ \ \\ b - (k-y) \cdot \ 9=x \ \\ 222 - (30-y) \cdot \ 9=x \ \\ x=9n+6 \ \\ y=n+6 \ \\ n >=2 \ \\ 9n+6 <=222 \ \\ n<=24 \ \\ 2<=n <=24 \ \\ n=5 \ \\ x=9 \cdot \ n+6=9 \cdot \ 5+6=51$
$y=n+6=5+6=11$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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