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?
  
== 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) ==
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Call the number <math>\overline{abcd}</math>. Then <math>a+b=c+d</math>. Set <math>a+b=x</math>.
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Clearly, <math>0\le x \le18</math>.
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If <math>x=0</math>: <math>0000</math> is not acceptable.
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If <math>x=1</math>: The only case is <math>1001</math> or <math>1010</math>. 2 choices.
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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.
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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.
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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>.
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If <math>x=7</math>: <math>7*8=56</math>.
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If <math>x=8</math>: <math>8*9=72</math>.
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If <math>x=9</math>: <math>9*10=90</math>.
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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>.
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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>.
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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>.
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~hastapasta
  
 
== See also ==
 
== See also ==

Revision as of 12:57, 4 May 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, $0\le x \le18$.

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

See also

2003 AIME I (ProblemsAnswer KeyResources)
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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