Asked by Ben
Let d(n) equal the number of positive divisors of the integer n.
Find d(d(p^{p-1})) where p is any prime number.
Find d(d(p^{p-1})) where p is any prime number.
Answers
Answered by
oobleck
check a few primes
d(3^2) = 3. 1,3,9
d(5^4) = 5. 1,5,25,125,625
I expect it will not be hard to prove that d(p^(p-1)) = p
So now we're down to d(p) where p is prime...
d(3^2) = 3. 1,3,9
d(5^4) = 5. 1,5,25,125,625
I expect it will not be hard to prove that d(p^(p-1)) = p
So now we're down to d(p) where p is prime...
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