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A328864 For any three-digit number k = hdu, f(k) = (h+d+u) + (h*d+d*u+u*h) + (h*d*u). This sequence consists of the numbers k for which the ratio k/f(k) is an integer.

%I A328864 #58 Feb 16 2025 08:33:58
%S A328864 100,114,115,120,121,190,199,200,207,208,210,221,260,290,299,300,301,
%T A328864 304,330,390,399,400,420,441,448,490,499,500,572,573,590,599,600,620,
%U A328864 624,625,690,699,700,705,790,799,800,806,880,890,899,900,990,999
%N A328864 For any three-digit number k = hdu, f(k) = (h+d+u) + (h*d+d*u+u*h) + (h*d*u). This sequence consists of the numbers k for which the ratio k/f(k) is an integer.
%C A328864 The idea of this sequence comes from the 1st problem of the 30th British Mathematical Olympiad in 1994 [see link BMO].
%C A328864 This sequence is finite with 50 terms.
%C A328864 The values of k/f(k) obtained are 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 13, 15, 16, 22, 24, 30, 31, 42, 43, 100.
%C A328864 Three particular subsequences:
%C A328864 k/f(k) = 1 for k = 199, 299, 399, 499, 599, 699, 799, 899, 999 (answer to part (ii) of the BMO problem).
%C A328864 k/f(k) = 10 for k = 190, 290, 390, 490, 590, 690, 790, 890, 990.
%C A328864 k/f(k) = 100 for k = 100, 200, 300, 400, 500, 600, 700, 800, 900.
%C A328864 Other definition: three-digit numbers k = hdu such as k/(e_1(h,d,u) + e_2(h,d,u) + e_3(h,d,u)) is an integer, where e_1, e_2, e_3 are the elementary symmetric polynomials in 3 variables.
%C A328864 Remark: When k has two digits du, the numbers that are divisible by (e_1(d,u) + e_2(d,u)) = (d+u) + (d*u) are the first 19 terms of A038366.
%D A328864 A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Pb 1 pp. 55 and 99-100 (1994)
%H A328864 British Mathematical Olympiad, <a href="https://bmos.ukmt.org.uk/home/bmo1-1994.pdf">1994 - Problem 1</a>.
%H A328864 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SymmetricPolynomial.html">Symmetric polynomial</a>
%H A328864 Wikipedia, <a href="https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial">Elementary symmetric polynomial</a>.
%e A328864 For k = 625, f(k) = 6+2+5 + 6*2+2*5+6*5 + 6*2*5 = 13 + 52 + 60 =  125 and 625/125 = 5, hence, 625 is a term, and 5 is the solution to part (i) of the BMO problem.
%p A328864 for i from 1 to 9 do
%p A328864 for j from 0 to 9 do
%p A328864 for k from 0 to 9 do
%p A328864      n := 100*i + 10*j + k ;
%p A328864      m := i + j + k + i*j + j*k + k*i + i*j*k ;
%p A328864      if n/m = floor(n/m) then print(n,m,n/m) ; end if ;
%p A328864 end do ;
%p A328864 end do ;
%p A328864 end do ;
%t A328864 Select[Range[100, 999], ({h,d,u} = IntegerDigits@ #; IntegerQ[# / (d + u + d u + (1 + d) h (1 + u))]) &] (* _Giovanni Resta_, Oct 29 2019 *)
%Y A328864 Cf. A038366 (similar, with 2 digits, the first 19 terms).
%Y A328864 Cf. A005349 (Niven numbers), A007602 (Zuckerman numbers).
%K A328864 nonn,fini,full,base
%O A328864 1,1
%A A328864 _Bernard Schott_, Oct 29 2019