2024 AIME II Problems/Problem 3

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

Solution 2

Let this be the table:

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

Then, $100(a+d)+10(b+e)+(c+f)=999$ and $10(a+b+c)+(d+e+f)=99$ . We look at the first equation. Since $c$ and $f$ are digits, $c<9$ and $f<9$ , so $c+f<18$ , which means $c+f=9$ for the last digit to be $9$ . Similarly, $b+e=9$ and $a+d=9$ .

Page

Toolbox

Search

2024 AIME II Problems/Problem 3

Problem

Solution 1

Solution 2