Question

An integer between $1000$ and $9999,$ inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there?

Answer

Solution:

Group the balanced integers by the common sum $s$ of each digit pair, where $1 \le s \le 18.$ For $s \le 9,$ the leftmost pair (first digit at least $1$ ) can be formed in $s$ ways and the rightmost pair in $s + 1$ ways. For $s \ge 10,$ both digits of each pair must be at least $s - 9,$ giving $19 - s$ ways for each pair.

The total count is $\sum_{s=1}^{9} s(s+1) + \sum_{s=10}^{18} (19 - s)^2 = \sum_{s=1}^{9} (s^2 + s) + \sum_{k=1}^{9} k^2 = 2 \cdot 285 + 45 = 615.$

2003 AIME I Problem 9

Solution:

Problem 9 in Other Years