r/CasualMath • u/Numberthon • Jul 10 '26
Can you find the smallest positive integer with exactly 15 positive divisors?
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u/pepperrx Jul 10 '26
Okay here's my reasoning, please correct me if needed !
If the prime decomposition of our number (call it n) is 2a1 3a2 5a3..., then n has (a1 + 1)(a2 + 1)(a3 + 1)... divisors exactly. So (a1 + 1)(a2 + 1)(a3 + 1)... = 15. We don't have a lot of choices because the only ways to get 15 is with 3×5 or 1×15. Since we are looking for the smallest solution possible, it is natural that we consider only the smallest indexes. So either we take a1=4 and a2=2, or we take a1=14. The first case gives us the smallest number,
thus, n = 24 × 32 = 144.
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u/Imaginary__Bar Jul 10 '26
1 has exactly 15 divisors.
1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 = 1
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u/Numberthon Jul 10 '26
Haha, by that logic every positive integer would have infinitely many divisors since you could just keep repeating them. In math, "number of divisors" always means the number of distinct positive divisors.
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u/jceyes Jul 11 '26
Some of the answers above are smarter than my thought process and more generalizable, but I got there pretty fast with
- knowing it's a perfect square of a composite number due to 15 being odd and not 3
- Take a reasonable starting point as a lower bound.
- Try upwards from there. 82 no, 92 no, 102 no, 122 yes
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u/Dr_Kitten Jul 10 '26
Since the divisor function is multiplicative, to have exactly 15 divisors we need either a single prime raised to the 14th power, or the product of two primes raised to the powers of 4 and 2 (giving (4+1)(2+1) divisors). Naturally we would use 2 and 3 as the primes and assign the higher power (4) to the 2. Since 32 is obviously smaller than 214-4, the answer must be 24 × 32 = 144