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Difference between revisions of "2003 AIME I Problems/Problem 9"
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An [[integer]] between <math>1000</math> and <math>9999</math>, inclusive, is called ''balanced'' if the sum of its two leftmost [[digit]]s equals the sum of its two rightmost digits. How many balanced integers are there?
 
An [[integer]] between <math>1000</math> and <math>9999</math>, inclusive, is called ''balanced'' if the sum of its two leftmost [[digit]]s equals the sum of its two rightmost digits. How many balanced integers are there?
  
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== Solution ==
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== Solution 1==
 
If the common sum of the first two and last two digits is <math>n</math>, such that <math>1 \leq n \leq 9</math>, there are <math>n</math> choices for the first two digits and <math>n + 1</math> choices for the second two digits (since zero may not be the first digit).  This gives <math>\sum_{n = 1}^9 n(n + 1) = 330</math> balanced numbers.  If the common sum of the first two and last two digits is <math>n</math>, such that <math>10 \leq n \leq 18</math>, there are <math>19 - n</math> choices for both pairs.  This gives <math>\sum_{n = 10}^{18} (19 - n)^2 = \sum_{n = 1}^9 n^2 = 285</math> balanced numbers.  Thus, there are in total <math>330 + 285 = \boxed{615}</math> balanced numbers.
 
If the common sum of the first two and last two digits is <math>n</math>, such that <math>1 \leq n \leq 9</math>, there are <math>n</math> choices for the first two digits and <math>n + 1</math> choices for the second two digits (since zero may not be the first digit).  This gives <math>\sum_{n = 1}^9 n(n + 1) = 330</math> balanced numbers.  If the common sum of the first two and last two digits is <math>n</math>, such that <math>10 \leq n \leq 18</math>, there are <math>19 - n</math> choices for both pairs.  This gives <math>\sum_{n = 10}^{18} (19 - n)^2 = \sum_{n = 1}^9 n^2 = 285</math> balanced numbers.  Thus, there are in total <math>330 + 285 = \boxed{615}</math> balanced numbers.
  
 
Both summations may be calculated using the formula for the [[perfect square|sum of consecutive squares]], namely <math>\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}</math>.
 
Both summations may be calculated using the formula for the [[perfect square|sum of consecutive squares]], namely <math>\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}</math>.
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== Solution 2 (Painful Casework) ==
 
== Solution 2 (Painful Casework) ==

Revision as of 17:35, 29 April 2022

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, $2\le x \le18$.

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

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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