Choose 18 distinct 2-digit positive integers such that:
their sum is a 3-digit number with 0 as the middle digit, and
if you delete the 0 from that sum, you get a 2-digit number which equals the grand sum of all 36 digits.
Is the solution unique?
Let T be the sum of the 18 tens digits and U be the sum of the 18 ones digits.
Then let the 3-digit sum be A0B and then two digit sum is AB. Expressed algebraically, the 3-digit sum can be written as 100*A+B and the two digit sum can be written as 10*A+B.
Then the two sums are 10*T+U = 100*A+B and T+U = 10*A+B. Now solving for U in terms of A and B gives the very simple result U=B. But B is a digit. Then U is at most 9.
However we need 18 different digits to fill out the members of U that total to at most 9. There can be a maximum of nine 0s among the members of U, so the other nine members of U must be 1s to not exceed the upper bound of U=9.
So then we need to make a set of 18 distinct two digit numbers such that all the unit digits are 0 or 1; but there exactly 18 such numbers: the set {10,11,20,21,...90,91}. The sum of this set is 909, and the sum of digits is 99. This is the only solution.
Solution
