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1) If ⌊n/k⌋ divides n for 1 ≤ k ≤ n then n is divisible by any integer 2 ≤ a

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Problem 11747

(American Mathematical Monthly, Vol.121, January 2014) Proposed by J. C. Lagarias (USA).

Determine alln ∈ N such that ⌊n/k⌋ divides n for 1 ≤ k ≤ n. Similarly, determine all n ∈ N such that⌈n/k⌉ divides n for 1 ≤ k ≤ n.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We will show that

i) ⌊n/k⌋ divides n for 1 ≤ k ≤ n iff n ∈ {1, 2, 3, 4, 6, 8, 12, 24}, ii) ⌈n/k⌉ divides n for 1 ≤ k ≤ n iff n ∈ {1, 2, 3, 4, 6, 8, 12}.

We divide the proof in four steps.

1) If ⌊n/k⌋ divides n for 1 ≤ k ≤ n then n is divisible by any integer 2 ≤ a ≤√ n − 1.

In fact, given 2 ≤ a ≤√ n − 1,

a = ⌊n/k⌋ ⇐⇒ a ≤n

k < a + 1, ⇐⇒ k ∈

 n

a + 1,n a



⊂ [1, n].

Since n

a− n

a + 1 = n

a(a + 1) ≥ n

√n(√

n − 1)> 1 it follows that there exists an integer k ∈ [1, n] such that a = ⌊n/k⌋.

2) If ⌈n/k⌉ divides n for 1 ≤ k ≤ n then n is divisible by any integer 2 ≤ a ≤√ n − 1.

In fact, given 2 ≤ a ≤√ n − 1,

a = ⌈n/k⌉ ⇐⇒ a − 1 < n

k ≤ a, ⇐⇒ k ∈ n a, n

a − 1



⊂ [1, n].

Since n

a − 1−n

a = n

a(a − 1) ≥ n

(√

n − 1)(√

n − 2) > 1 it follows that there exists an integer k ∈ [1, n] such that a = ⌈n/k⌉.

3) If n is divisible by any integer 2 ≤ a ≤√

n − 1 then n < 64.

Let a = ⌊√

n⌋ − 1, then gcd(a, a − 1) = 1, a(a − 1) divides n and n = qa(a − 1) for some integer q > 1 (because a(a − 1) < n). Hence

(a + 2)2≥ n = qa(a − 1) ≥ 2a(a − 1) ⇐⇒ a2− 6a − 4 ≤ 0 ⇐⇒ (a − 3)2≤ 13, and a ≤ 6, that is ⌊√

n⌋ = a + 1 ≤ 7, or n < 64.

4) It is easy to verify that our statement holds for n < 64.



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Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.. We will show a more

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