Question

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Answer 1

Step-by-step explanation:

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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Hope this helps (((:

Answer 2

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.)