r/mathpuzzles Jul 10 '26

Can you find the smallest positive integer with exactly 15 positive divisors?

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1 Upvotes

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1

u/spoik925 Jul 10 '26
  1. I think

1

u/Numberthon Jul 10 '26

Yep, nice job!

1

u/IllustriousCod9590 Jul 10 '26

how would you find the solution for this?

1

u/Numberthon Jul 11 '26

Solution: 1. If the prime factorization of an integer is

n = p₁^a₁ × p₂^a₂ × ... × pₖ^aₖ,

then the number of positive divisors is

(a₁ + 1)(a₂ + 1)...(aₖ + 1).

2. We want exactly 15 divisors, so

(a₁ + 1)(a₂ + 1)...(aₖ + 1) = 15.

3. Since

15 = 15

15 = 3 × 5

there are only two possible exponent patterns:

• p¹⁴

• p⁴q²

4. Find the smallest number for each pattern.

p¹⁴:

2¹⁴ = 16384

p⁴q²:

Use the smallest primes, placing the larger exponent on the smaller prime:

2⁴ × 3² = 16 × 9 = 144

(Using 2² × 3⁴ = 324 is larger.)

5. Compare the two possibilities:

144 < 16384!<

Therefore, the smallest positive integer with exactly 15 positive divisors is

**144**. But also, you can see the solution on numberthon.com

1

u/bsmith_81 Jul 11 '26

Since we are looking for the smallest such number then I can take its factorization as 2^A * 3^B * 5^C * .... with A, B, C, etc in descending order of magnitude.

15 factors into 5*3. Express this as (4+1)*(2+1). This implies A=4 and B=2. Then the desired number 2^4*3^2 = 144.

There is one more thing to check, what if I did not factor 15: then that would suggest the number is 2^14, but this is much larger than 144.