A335491 a(n) is the smallest number m with exactly n divisors whose last digit equals the last digit of m.

1, 11, 40, 60, 160, 120, 640, 240, 360, 480, 8064, 600, 18144, 1920, 1440, 1200, 72576, 1800, 52416, 2400, 5760, 30720, 183456, 3600, 12960, 122880, 9000, 9600, 602784, 7200, 445536, 8400, 92160, 798336, 51840, 12600, 2159136, 576576, 368640, 16800, 2935296, 28800

Offset: 1

Author

Bernard Schott, Jun 11 2020

Keywords

• nonn
• base

Comments

a(n) exists for any n >= 1. Indeed, the number 5*2^n (see A020714), n >= 1, has exactly n divisors (5*2^1, 5*2^2, ..., 5*2^n), with the last digit 0. - Marius A. Burtea, Jun 12 2020

It's always the case when a(n) ends in 0 then a(n) = 10 * A005179(n). Proof: Let v be the list of divisors of a(n) that end in 0. We then have |v| = n and lcm(v) = a(n) as a(n) is in v and all other terms in v divide a(n). We then have lcm(v)/10 = a(n)/10 where a(n)/10 has exactly n divisors. The least positive integer that has exactly n divisors is A005179(n). - Bernard Schott and David A. Corneth, Jun 12 2020

For some p_i^e_i||a(n) and p_j^e_j||a(n) where p_i and p_j are primes such that p_i < p_j, 10 | (p_j - p_i) and t^k||u denotes t^k|u but t^(k + 1) doesn't divide u i.e. gcd(t, u/t^k) = 1 denotes then e_i >= e_j. For example k * 11^2 * 31^3 where gcd(k, 11*31) = 1 can't be a term as the multiplicity of 11 is less than the multiplicity of 31. - David A. Corneth, Jun 12 2020

Examples

			Of the twelve divisors of 60, four have their last digit equals to the last digit of 60: 10, 20, 30 and 60, and there is no smaller number k with four divisors whose last digit equals the last digit of k, hence a(4) = 60.
		

Links

• Project Euler, Problem 474: Last digits of divisors

Crossrefs

Cf. A000005, A005179, A020714, A330348.

Similar with: A333456 (Niven numbers), A335038 (Zuckerman numbers).

Programs

Magma

a:=[]; for n in [1..30] do k:=1; while #[d:d in Divisors(k)|k mod 10 eq d mod 10] ne n do k:=k+1; end while; Append(~a,k); end for; a; // Marius A. Burtea, Jun 12 2020

Mathematica

d[n_] := DivisorSum[n, 1 &, Mod[# - n, 10] == 0 &]; mx = 20; c = 0; n = 1; s = Table[0, {mx}]; While[c < mx, i = d[n]; If[i <= mx && s[[i]] == 0, c++; s[[i]] = n]; n++]; s (* Amiram Eldar, Jun 12 2020 *)

PARI

f(n) = my(u=n%10); sumdiv(n, d, (d%10) == u);
a(n) = my(k=1); while(f(k) != n, k++); k; \\ Michel Marcus, Jun 12 2020

Extensions

Corrected and extended by Marius A. Burtea, Jun 12 2020