Arrange the digits 1 to 9 to form three 3-digit numbers such that the sum of the two smaller numbers equals the larger.
arithmetic/digits/three.digit.sum New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
Arrange the digits 1 to 9 to form three 3-digit numbers such that the
sum of the two smaller numbers equals the larger. How many solutions
are there?
Tear up a piece of paper into 9 pieces. Write the digits 1 to 9 on a
piece of paper, one digit per piece.
Now rearrange the bits of paper to form three 3-digit numbers, such that
the sum of the two smaller numbers equals the larger. Here is a failed
attempt.
2 7 3
+ 6 1 5
----------
9 4 8
There is more than one solution. How many solutions are there
altogether?
-- rex@bertha.ucq.edu.au, September 1994
Show solution
A start in finding all solutions is to consider the addition done (mod 9). Since any number equals the sum of its base 10 digits (mod 9), let the sum of the 6 digits appearing in the two addends be S. Then we must have that S = 1+2+...+9 - S (mod 9), or 2*S = (1+9)*9/2 (mod 9) ==> 2*S = 45 (mod 9) ==> S = 0 (mod 9). This implies that the summand must be divisible by 9, thus the final digit is fixed after choosing the first two. For each solution, you can also clearly exchange digits among the addends, so the final total will be divisible by 4 (you can also use this observation to speed up a computer search). -- Chris Long <clong@cnj.digex.net>, September 1994 [ Editor: no count of the solutions was found in the collected material. ]