Elgin Johnston, Problems Editor
Department of Mathematics
Iowa State University
Ames, IA 50011
1632. Proposed by Erwin Just (Emeritius), Bronx Community College New York, NY.
Prove that there are an infinite number of integers n for which there exists a set
of n distinct odd integers such that each member of the set divides the sum of all the
members of the set.
Proof.
1.)
Note that 945 = 1 + 9 + 21 + 27 + 35 + 45 + 63 + 105 + 135 + 189 + 315 and that
each term is a divisor of 945. Label these divisors, in order of increasing
magnitude, d1 , d2 , … , d11 .
2.)
Claim: for m ≥ 2,
945
m-1
m
= 945 +
11
945 j dk
j=1
k=2
The proof is by induction.
945
2
= 945( 1 + 9 + 21 + 27 + 35 + 45 + 63 + 105 + 135 + 189 + 315 )
= 945 + 945( 9 + 21 + 27 + 35 + 45 + 63 + 105 + 135 + 189 + 315 ).
945
m+1
= 945( 945

= 945 945 +

m
)

945 dk
j=1
k=2 
m-1

11
j
m
11
= 945 + 945 di
+
j=2
k=2
m
= 945 +
11
945 dk
j=1
Allen J. Mauney
j
11
945 j dk
k=2
, as desired.
k=2
1215 Kaiser Road NW.
Olympia, WA 98502
3.)
All terms of the form 945
odd, divisors of 945
m
corresponding to 945
m
j
dk
, 1 ≤ j ≤ m – 1 and 1 ≤ k ≤ 11, are clearly distinct,
and, by construction, they sum to 945
m
. Thus,
, m ≥ 2, there is a set of n = 1 + 10( m – 1 ) numbers that
satisfy all of the conditions of the problem.
QED
Allen J. Mauney
1215 Kaiser Road NW.
Olympia, WA 98502