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# Number Theory

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Let \$m\$ be a positive integer.  If \$m\$ has exactly \$18\$ positive divisors, then how many positive divisors does \$m^2\$ have?

Jun 24, 2024

#1
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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

\(32400 =2 ^ 4 ×3 ^ 4 ×5 ^ 2\)

\((4+1)(4+1)(2+1)=5×5×3=75\)

Thanks! :)

Jun 24, 2024
edited by NotThatSmart  Jun 24, 2024
#2
+129725
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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

Jun 24, 2024