SOLUTION: find the smallest number which leaves remainder 8 and 12 when divided by 28 & 32, respectively.

SOLUTION: find the smallest number which leaves remainder 8 and 12 when divided by 28 & 32, respectively.

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Question 288318: find the smallest number which leaves remainder 8 and 12 when divided by 28 & 32, respectively.
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
28x+8=z
32y+12=z
{4(7x+2)= z,4(8y+3)=z}
7x+2=8y+3
7x=8y+1
7x-1=8y
(7x-1)/8=y
x=(8y+1)/7
x=8n-1, y=7n-1
let n=1
x=8-1
y=7-1
x=7 y=6
28*7+8=32*6+12=204
check
204-8=196/28=7
204-12=192/32=6
ok
Answer 204

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