SOLUTION: A question on an assignment I have is "Provide an algebraic proof of the fact that any three-digit number of the form 100a + 10b + a is divisible by 7 when a + b = 7." I am having

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Question 1003366: A question on an assignment I have is "Provide an algebraic proof of the fact that any three-digit number of the form 100a + 10b + a is divisible by 7 when a + b = 7." I am having a lot of trouble figuring this one out. I understand the divisibility rule for 7 but can't figure out how to put it into an algebraic expression.
Found 2 solutions by Edwin McCravy, richard1234:
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
Prove that 100a + 10b + a is divisible by 7 when a + b = 7
Since a + b = 7, b = 7 - a.  Substituting that in

100a + 10b + a  gives

100a + 10(7 - a) + a

100a + 70 - 10a + a

91a + 70

7(13a+10)

That has factor 7, and so it is divisible by 7.

Edwn
Answer by richard1234(7193)   (Show Source): You can put this solution on YOUR website!


Since 98a+7b is always divisible by 7, we only need to check if 3a + 3b is divisible by 7. Since 3 and 7 are relatively prime, this occurs if and only if a+b is divisible by 7.
More formally, we say that .
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