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Difference between revisions of "2003 AIME I Problems/Problem 9"

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== Problem ==
 
== Problem ==
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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?
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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?
  
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== Solution ==
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== Solution 1==
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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>.
 +
 
 +
== Solution 2 (Painful Casework) ==
 +
 
 +
Call the number <math>\overline{abcd}</math>. Then <math>a+b=c+d</math>. Set <math>a+b=x</math>.
 +
 
 +
Clearly, <math>0\le x \le18</math>.
 +
 
 +
If <math>x=0</math>: <math>0000</math> is not acceptable.
 +
 
 +
If <math>x=1</math>: The only case is <math>1001</math> or <math>1010</math>. 2 choices.
 +
 
 +
If <math>x=2</math>: then since <math>a\neq0</math>, <math>a=1=b</math> or <math>a=2, b=0</math>. There are 3 choices for <math>(c,d)</math>: <math>(2,0), (0, 2), (1, 1)</math>. <math>2*3=6</math> here.
 +
 
 +
If <math>x=3</math>: Clearly, <math>a\neq b</math> because if so, the sum will be even, not odd. Counting <math>(a,b)=(3,0)</math>, we have <math>4</math> choices. Subtracting that, we have <math>3</math> choices. Since it doesn't matter whether <math>c=0</math> or <math>d=0</math>, we have 4 choices for <math>(c,d)</math>. So <math>3*4=12</math> here.
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 +
If <math>x=4</math>: Continue as above. <math>4</math> choices for <math>(a,b)</math>. <math>5</math> choices for <math>(c,d)</math>. <math>4*5=20</math> here.
 +
 
 +
If <math>x=5</math>: You get the point. <math>5*6=30</math>.
 +
 
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If <math>x=6</math>: <math>6*7=42</math>.
 +
 
 +
If <math>x=7</math>: <math>7*8=56</math>.
 +
 
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If <math>x=8</math>: <math>8*9=72</math>.
 +
 
 +
If <math>x=9</math>: <math>9*10=90</math>.
 +
 
 +
Now we need to be careful because if <math>x=10</math>, <math>(c,d)=(0,10)</math> is not valid. However, we don't have to worry about <math>a\neq0</math>.
 +
 
 +
If <math>x=10</math>: <math>(a,b)=(1,9), (2, 8), ..., (9, 1)</math>. Same thing for <math>(c,d)</math>. <math>9*9=81</math>.
 +
 
 +
If <math>x=11</math>: We start at <math>(a,b)= (2,9)</math>. So <math>8*8</math>.
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Continue this pattern until <math>x=18: 1*1=1</math>. Add everything up: we have <math>\boxed{615}</math>.
 +
 
 +
~hastapasta
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 +
== 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 <math>a</math> and <math>b</math> sum to <math>0</math>, we have <math>1</math> possibility: <math>(a,b) = (0,0)</math>
 +
 
 +
If <math>a+b = 1</math>, we have <math>2</math> possibilities: <math>(a,b) = (0,1)</math> and <math>(a,b) = (1,0)</math>
 +
 
 +
<math>a+b = 2</math>: <math>(0,2)</math>, <math>(2,0)</math>, and <math>(1,1)</math> are the only <math>3</math> possibilities.
 +
 
 +
We notice a pattern: for all <math>k \leq 9</math>, there are <math>k+1</math> ordered pairs of digits <math>(a,b)</math> such that <math>a+b = k</math>. Then, testing for <math>10 \leq k \leq 18</math>, we notice that there are <math>(18 - k) + 1</math> ordered pairs with a sum of <math>k</math>.
 +
 
 +
So the number of ways to pick <math>4</math> digits satisfying the constraint is
 +
<cmath>(1)(1) + (2)(2) \dots (10)(10) + (9)(9) + \dots + (1)(1) = \frac{(10)(11)(21)}{6} + \frac{(9)(10)(19)}{6} = 670</cmath>
 +
 
 +
We have to subtract out the cases where the first digit is <math>0</math>, which is <math>1+2+\dots+10  = 55</math>
 +
 
 +
So our answer is <math>670-55 = \boxed{615}</math>
 +
 
 +
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 <math>10</math>-sided dice.
 +
 
 +
~jd9
  
 
== See also ==
 
== See also ==
−
* [[2003 AIME I Problems]]
+
{{AIME box|year=2003|n=I|num-b=8|num-a=10}}
 +
 
 +
[[Category: Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 10:55, 10 September 2023

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

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AIME Problems and Solutions

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