ANSWERS: 2

8^n is divisible by 8, obviously, so we only have to prove that 9^n1 is divisible by 8. This can be proven by recurrence. It is valid for n=1 because 91=8. (it would even be valid for n=0 because 11=0 which is divisible by 8). Now let's assume that it is valid for (n1): 9^(n1)1 is divisible by 8. We can write: 9^n1=(8+1)*9^(n1)1=8*9^(n1) + 9^(n1) 1. Now the first term 8*9^(n1) is obviously divisible by 8 and the rest 9^(n1) 1 is divisible by 8 because the statement is valid for (n1).

1262017 Let n=2. Then 81 + 8 = 89 and your premise is disproved.


iwnitJewels Vern: if it were that way, it should be written 9^n+8^(n1). I read the expression as (9^n) + (8^n)  1. This last expression fulfills the premise in your example: 81 + 64 1 = 81 + 63 = 9*(9 +7) .

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