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Difference between revisions of "2003 AIME I Problems/Problem 9"
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Call the number <math>\overline{abcd}</math>. Then <math>a+b=c+d</math>. Set <math>a+b=x</math>.
 
Call the number <math>\overline{abcd}</math>. Then <math>a+b=c+d</math>. Set <math>a+b=x</math>.
  
−
Clearly, <math>2\lex\le18</math>.
+
Clearly, <math>2\le x \le18</math>.
  
 
If <math>x=1</math>: The only case is <math>1001</math> or <math>1010</math>. 2 choices.
 
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>2*3=6<math> here.
+
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\neqb<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.
+
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.
  
−
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=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>.
+
If <math>x=5</math>: You get the point. <math>5*6=30</math>.
  
−
If </math>x=6<math>: </math>6*7=42<math>.
+
If <math>x=6</math>: <math>6*7=42</math>.
  
−
If </math>x=7<math>: </math>7*8=56<math>.
+
If <math>x=7</math>: <math>7*8=56</math>.
  
−
If </math>x=8<math>: </math>8*9=72<math>.
+
If <math>x=8</math>: <math>8*9=72</math>.
  
−
If </math>x=9<math>: </math>9*10=90<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>.
+
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=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>.
+
If <math>x=11</math>: We start at <math>(a,b)= (2,9)</math>. So <math>8*8</math>.
  
−
Continue this pattern until </math>x=18: 1*1=1<math>. Add everything up: we have </math>\boxed{615}$.
+
Continue this pattern until <math>x=18: 1*1=1</math>. Add everything up: we have <math>\boxed{615}</math>.
  
 
~hastapasta
 
~hastapasta

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

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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