Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
(→‎Solution 2 (Painful Casework))
(10 intermediate revisions by 5 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== 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?
+
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 ==
+
== 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.
 +
 
 +
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.
 +
 
 +
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=6</math>: <math>6*7=42</math>.
 +
 
 +
If <math>x=7</math>: <math>7*8=56</math>.
 +
 
 +
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>.
 +
 
 +
Continue this pattern until <math>x=18: 1*1=1</math>. Add everything up: we have <math>\boxed{615}</math>.
 +
 
 +
~hastapasta
  
 
== 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}}

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_I_Problems/Problem_9&oldid=173773"