2003 AIME I Problem 9

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

Answer: 615

Concepts:digits casework sum of first n squares

Difficulty rating: 2430

Small hint:

Group by the common digit-pair sum

s;

count leftmost pairs (first digit nonzero) and rightmost pairs separately

Big hint:

For

s \le 9

the counts are

s

and

s + 1;

for

s \ge 10

both equal

19 - s.

Sum the products.

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

\begin{aligned} &\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. \end{aligned}