ANSWERS: 2
  • 8^n is divisible by 8, obviously, so we only have to prove that 9^n-1 is divisible by 8. This can be proven by recurrence. It is valid for n=1 because 9-1=8. (it would even be valid for n=0 because 1-1=0 which is divisible by 8). Now let's assume that it is valid for (n-1): 9^(n-1)-1 is divisible by 8. We can write: 9^n-1=(8+1)*9^(n-1)-1=8*9^(n-1) + 9^(n-1) -1. Now the first term 8*9^(n-1) is obviously divisible by 8 and the rest 9^(n-1) -1 is divisible by 8 because the statement is valid for (n-1).
  • 1-26-2017 Let n=2. Then 81 + 8 = 89 and your premise is disproved.
    • iwnit
      Jewels Vern: if it were that way, it should be written 9^n+8^(n-1). 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) .

Copyright 2020, Wired Ivy, LLC

Answerbag | Terms of Service | Privacy Policy