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 1)
 
(2 intermediate revisions by one other user not shown)
Line 30: Line 30:
 
</math>
 
</math>
  
−
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) = 99</math>, or <math>9(a+b+c+3) = 99</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
−
wouldn’t a+b+c be 6?
 
  
 
==Video Solution==
 
==Video Solution==
Line 42: Line 41:
 
{{AIME box|year=2024|num-b=2|num-a=4|n=II}}
 
{{AIME box|year=2024|num-b=2|num-a=4|n=II}}
 
   
 
   
−
[[Category:]]
+
[[Category:Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:08, 5 March 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 $10(a+b+c) + (9-a+9-b+9-c) = 99$, or $9(a+b+c+3) = 99$, 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

Video Solution

https://youtu.be/nKRfXAHaQvA

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2024 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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=2024_AIME_II_Problems/Problem_3&oldid=216492"