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2003 AIME I Problems/Problem 9

Revision as of 17:24, 29 April 2022 by Hastapasta (talk | contribs)

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

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, $2\lex\le18$ (Error compiling LaTeX. Unknown error_msg).

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$ (Error compiling LaTeX. Unknown error_msg)x=3$: Clearly,$a\neqb$because if so, the sum will be even, not odd. Counting$(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$ (Error compiling LaTeX. Unknown error_msg)x=4$: Continue as above.$4$choices for$(a,b)$.$5$choices for$(c,d)$.$4*5=20$here.

If$ (Error compiling LaTeX. Unknown error_msg)x=5$: You get the point.$5*6=30$.

If$ (Error compiling LaTeX. Unknown error_msg)x=6$:$6*7=42$.

If$ (Error compiling LaTeX. Unknown error_msg)x=7$:$7*8=56$.

If$ (Error compiling LaTeX. Unknown error_msg)x=8$:$8*9=72$.

If$ (Error compiling LaTeX. Unknown error_msg)x=9$:$9*10=90$.

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

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

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

Continue this pattern until$ (Error compiling LaTeX. Unknown error_msg)x=18: 1*1=1$. Add everything up: we have$\boxed{615}$.

~hastapasta

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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