3^k is a power of 3 and the factors of this number are 1, 3 and the higher powers of 3 up to 3^(k-1). For a number to be divisible by 40, the number should have three 2s and 5 as prime factors. Since 3^k does not have these prime factors,it is not divisble by 40 for any value of k!]]>

3^1 = 3

3^4 = 81

3^13 = 1594323

3^17 = 129140163

Since none of these divide into 40 without a remainder, I'd say the answer is e.

]]>a. 1

b. 4

c. 13

d. 17

e. none of these

I know a can't be right as 40 is to even of a number to be that only.

but do I use the k like tha other problem with 1/2^3, 1/2^4...and so on? Or is there another one???