2024 AIME II Problems/Problem 3

Revision as of 16:19, 14 February 2024 by Iwowowl253 (talk | contribs) (Solution 1)

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 wouldn’t a+b+c be 6?

Video Solution

https://youtu.be/nKRfXAHaQvA

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

See also

2024 AIME II (ProblemsAnswer KeyResources)
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

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