SOLUTION: In what bases, b, does (b+7) divide into (9b+7) without any remainder?

Question 1110462: In what bases, b, does (b+7) divide into (9b+7) without any remainder?
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

We have to have equal to a whole number; and presumably the value(s) of b we are looking for are integers greater than or equal to 2.

Since 9 is a whole number, has to be a whole number also.

So we know b has to be a whole number greater than or equal to 2; and we know 56 has to be divisible by (b+7). The only divisors of 56 that are 9 or greater are 14, 28, and 56.

Since those divisors are the values of (b+7), the three bases for which (b+7) divides into (9b+7) without any remainder are 7, 21, and 49.

Check:
b=7: (9b+7)(b+7) = 70/14 = 5
b=21: (9b+7)/b+7) = 196/28 = 7
b=49: (9b+7)/b+7) = 448/56 = 8
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