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"

(→‎Solution 2 (Painful Casework))
 
(One intermediate revision by the same user not shown)
Line 42: Line 42:
  
 
~hastapasta
 
~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 <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 ==

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
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=198039"