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.

100, 114, 115, 120, 121, 190, 199, 200, 207, 208, 210, 221, 260, 290, 299, 300, 301, 304, 330, 390, 399, 400, 420, 441, 448, 490, 499, 500, 572, 573, 590, 599, 600, 620, 624, 625, 690, 699, 700, 705, 790, 799, 800, 806, 880, 890, 899, 900, 990, 999

Offset: 1

Author

Bernard Schott, Oct 29 2019

Keywords

• nonn
• fini
• full
• base

Comments

The idea of this sequence comes from the 1st problem of the 30th British Mathematical Olympiad in 1994 [see link BMO].

This sequence is finite with 50 terms.

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.

Three particular subsequences:

k/f(k) = 1 for k = 199, 299, 399, 499, 599, 699, 799, 899, 999 (answer to part (ii) of the BMO problem).

k/f(k) = 10 for k = 190, 290, 390, 490, 590, 690, 790, 890, 990.

k/f(k) = 100 for k = 100, 200, 300, 400, 500, 600, 700, 800, 900.

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.

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.

Examples

			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.
		

References

• 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)

Links

• British Mathematical Olympiad, 1994 - Problem 1.
• Eric Weisstein's World of Mathematics, Symmetric polynomial
• Wikipedia, Elementary symmetric polynomial.

Crossrefs

Cf. A038366 (similar, with 2 digits, the first 19 terms).

Cf. A005349 (Niven numbers), A007602 (Zuckerman numbers).

Programs

Maple

for i from 1 to 9 do
for j from 0 to 9 do
for k from 0 to 9 do
     n := 100*i + 10*j + k ;
     m := i + j + k + i*j + j*k + k*i + i*j*k ;
     if n/m = floor(n/m) then print(n,m,n/m) ; end if ;
end do ;
end do ;
end do ;

Mathematica

Select[Range[100, 999], ({h,d,u} = IntegerDigits@ #; IntegerQ[# / (d + u + d u + (1 + d) h (1 + u))]) &] (* Giovanni Resta, Oct 29 2019 *)