+1222 Let $m$ be a positive integer. If $m$ has exactly $18$ positive divisors, then how many positive divisors does $m^2$ have?
+1908 I'm not sure how to explain the prcoess, but the smallest number with 18 positive divisors is \(180\)
We have \(1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180.\)
Now, \(180^2 = 32400\)
You could also use prime factorization, by doing
\(32400 =2 ^ 4 ×3 ^ 4 ×5 ^ 2\)
\((4+1)(4+1)(2+1)=5×5×3=75\)
Thus, our answeris 75.
Thanks! :)
+129852 Depends on the factorization of m
m = 2 * 3^8 has (1 + 1) (8 + 1) = 18 divisors
m^2 = [ 2 * 3^8]^2 = 2^2 * 3^16 has ( 2 + 1) (16 + 1) = 51 divisors
But
m = 2^2 * 3^5 has (2+1)(5 + 1) = 18 divisors
m^2 = [ 2^2 * 3^5]^2 = 2^4 * 3^10 has (4 + 1) (10 + 1) = 55 divisors
And as NTS pointed out,
m = 2^2 * 3^2 * 5 has (2 + 1) (2 + 1) (1 + 1) =18 divisors
But
(2^2 * 3^2 * 5)^2 = 2^4 * 3^4 * 5^2 has ( 4 + 1 )(4 + 1) (2 + 1) = 75 divisors
