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Find all positive integers (n) such that n/d(n) is the prime number, Where d(n) is the number of divisors of n. For example, d(15)=4 {1,3,5,15}.
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Hey Vivek,
Thanks for the fun problem.
We decompose n = p₁^(a₁)p₂^(a₂)…pₙ^(aₙ) and assume that p₁=n/d(n). Then, we get:
p₁^(a₁ – 1)p₂^(a₂)…pₙ^(aₙ) = (a₁ + 1)(a₂ + 1)…(aₙ + 1).
Now, we note that pᵢ^(aᵢ) ≥ (aᵢ+1) with equality only if pᵢ = 2 and aᵢ = 1.
Then, we must have:
- a₁ = 3 ⇒ n = 8; (easy)
- a₁ = 2 ⇒ n = 9, 12, 18; (easy)
- a₁ = 1 ⇒ n = p₁m, p₁ > 2; Then, aᵢ + 1=2 and 2/n. p₂ = 2 and we get n = 8p, p > 2 or n = 12p, p > 3.
I just brushed over the case analysis and I hope I haven’t missed some solutions… Even though I suspect I have 😀
- a₁ = 3 ⇒ n = 8; (easy)
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Fantastic, Artur. 🙂 You didn’t miss a thing. A minor typo though – first one is 8, not 4. So, 8,9,12,18,8p,12p
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