The idea is to look at the prime factorization of a number.
n = (p₁)^(a₁) (p₂)^(a₂) ... (pk)^(ak).
Then the number of divisors for n equals (1 + a₁)(1 + a₂)...(1 + ak).
[The 1's account for possible powers of 0.]
So, for 10 divisors, we have two possibilities (due to 10 = 10 or 2 * 5):
(i) n = (p₁)^4 (p₂)^1 ==> (4+1)(1+2) = 10 divisors
(ii) n = (p₁)^9 ==> 5+1 = 10 divisors.
For the smallest (positive) integers, just keep the sizes of the primes as small as possible.
So, we check 2^4 * 3 = 48 versus 2^9 = 512.
So, the smallest positive integer with 10 divisors is 48.
(Check: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.)
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For the next one, use the next 'smallest pair of primes':
2^4 * 5 = 80 vs. 2 * 3^4 = 162 (2^9 is still too big!)
So, the next one is 80.
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Can you see what the next one is? (You may need to use 7...)