# 2003 AIME I Problems/Problem 9

## Problem

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?

## Solution 1

If the common sum of the first two and last two digits is $n$, such that $1 \leq n \leq 9$, there are $n$ choices for the first two digits and $n + 1$ choices for the second two digits (since zero may not be the first digit). This gives $\sum_{n = 1}^9 n(n + 1) = 330$ balanced numbers. If the common sum of the first two and last two digits is $n$, such that $10 \leq n \leq 18$, there are $19 - n$ choices for both pairs. This gives $\sum_{n = 10}^{18} (19 - n)^2 = \sum_{n = 1}^9 n^2 = 285$ balanced numbers. Thus, there are in total $330 + 285 = \boxed{615}$ balanced numbers.

Both summations may be calculated using the formula for the sum of consecutive squares, namely $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$.

## Solution 2 (Painful Casework)

Call the number $\overline{abcd}$. Then $a+b=c+d$. Set $a+b=x$.

Clearly, $0\le x \le18$.

If $x=0$: $0000$ is not acceptable.

If $x=1$: The only case is $1001$ or $1010$. 2 choices.

If $x=2$: then since $a\neq0$, $a=1=b$ or $a=2, b=0$. There are 3 choices for $(c,d)$: $(2,0), (0, 2), (1, 1)$. $2*3=6$ here.

If $x=3$: Clearly, $a\neq b$ because if so, the sum will be even, not odd. Counting $(a,b)=(3,0)$, we have $4$ choices. Subtracting that, we have $3$ choices. Since it doesn't matter whether $c=0$ or $d=0$, we have 4 choices for $(c,d)$. So $3*4=12$ here.

If $x=4$: Continue as above. $4$ choices for $(a,b)$. $5$ choices for $(c,d)$. $4*5=20$ here.

If $x=5$: You get the point. $5*6=30$.

If $x=6$: $6*7=42$.

If $x=7$: $7*8=56$.

If $x=8$: $8*9=72$.

If $x=9$: $9*10=90$.

Now we need to be careful because if $x=10$, $(c,d)=(0,10)$ is not valid. However, we don't have to worry about $a\neq0$.

If $x=10$: $(a,b)=(1,9), (2, 8), ..., (9, 1)$. Same thing for $(c,d)$. $9*9=81$.

If $x=11$: We start at $(a,b)= (2,9)$. So $8*8$.

Continue this pattern until $x=18: 1*1=1$. Add everything up: we have $\boxed{615}$.

~hastapasta

## Solution 3

We ignore the requirement that the first digit is non-zero, and do casework on the sum of the sum of the pairs of digits.

If two digits $a$ and $b$ sum to $0$, we have $1$ possibility: $(a,b) = (0,0)$

If $a+b = 1$, we have $2$ possibilities: $(a,b) = (0,1)$ and $(a,b) = (1,0)$ $a+b = 2$: $(0,2)$, $(2,0)$, and $(1,1)$ are the only $3$ possibilities.

We notice a pattern: for all $k \leq 9$, there are $k+1$ ordered pairs of digits $(a,b)$ such that $a+b = k$. Then, testing for $10 \leq k \leq 18$, we notice that there are $(18 - k) + 1$ ordered pairs with a sum of $k$.

So the number of ways to pick $4$ digits satisfying the constraint is $$(1)(1) + (2)(2) \dots (10)(10) + (9)(9) + \dots + (1)(1) = \frac{(10)(11)(21)}{6} + \frac{(9)(10)(19)}{6} = 670$$

We have to subtract out the cases where the first digit is $0$, which is $1+2+\dots+10 = 55$

So our answer is $670-55 = \boxed{615}$

This solution takes some inspiration from dice. In a way, we are counting the number of ways to roll a given number with a pair of $10$-sided dice.

~jd9

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 