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 "2024 AIME II Problems/Problem 3"

(→‎Solution 2)
(→‎Solution 2)
Line 31: Line 31:
  
 
We want <math>10(a+b+c) + (9-a+9-b+9-c) = 81</math>, or <math>9(a+b+c+3) = 81</math>, or <math>a+b+c=8</math>. Since zeroes are allowed, we just need to apply stars and bars on <math>a, b, c</math>, to get <math>\tbinom{8+3-1}{3-1} = \boxed{045}</math>. ~akliu
 
We want <math>10(a+b+c) + (9-a+9-b+9-c) = 81</math>, or <math>9(a+b+c+3) = 81</math>, or <math>a+b+c=8</math>. Since zeroes are allowed, we just need to apply stars and bars on <math>a, b, c</math>, to get <math>\tbinom{8+3-1}{3-1} = \boxed{045}</math>. ~akliu
−
 
−
==Solution 2==
 
−
Let this be the table:
 
−
 
−
<math>
 
−
\begin{array}{|c|c|c|} \hline
 
−
a & b & c \\ \hline
 
−
d & e & f\\ \hline
 
−
\end{array}
 
−
</math>
 
−
 
−
Then, <math>100(a+d)+10(b+e)+(c+f)=999</math> and <math>10(a+b+c)+(d+e+f)=99</math>. We look at the first equation. Since <math>c</math> and <math>f</math> are digits, <math>c<9</math> and <math>f<9</math>, so <math>c+f<18</math>, which means <math>c+f=9</math> for the last digit to be <math>9</math>. Similarly, <math>b+e=9</math> and <math>a+d=9</math>.
 

Revision as of 21:02, 8 February 2024

Problem

Find the number of ways to place a digit in each cell of a 2x3 grid so that the sum of the two numbers formed by reading left to right is $999$, and the sum of the three numbers formed by reading top to bottom is $99$. The grid below is an example of such an arrangement because $8+991=999$ and $9+9+81=99$.

$\begin{array}{|c|c|c|} \hline 0 & 0 & 8 \\ \hline 9 & 9 & 1\\ \hline \end{array}$

Solution 1

Consider this table:

$\begin{array}{|c|c|c|} \hline a & b & c \\ \hline d & e & f\\ \hline \end{array}$

We note that $c+f = 9$, because $c+f \leq 18$, meaning it never achieves a unit's digit sum of $9$ otherwise. Since no values are carried onto the next digit, this implies $b+e=9$ and $a+d=9$. We can then simplify our table into this:

$\begin{array}{|c|c|c|} \hline a & b & c \\ \hline 9-a & 9-b & 9-c \\ \hline \end{array}$

We want $10(a+b+c) + (9-a+9-b+9-c) = 81$, or $9(a+b+c+3) = 81$, or $a+b+c=8$. Since zeroes are allowed, we just need to apply stars and bars on $a, b, c$, to get $\tbinom{8+3-1}{3-1} = \boxed{045}$. ~akliu

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AIME_II_Problems/Problem_3&oldid=214972"