Difference between revisions of "2024 AIME II Problems/Problem 3"
R00tsofunity (talk | contribs) |
|||
(13 intermediate revisions by 6 users not shown) | |||
Line 3: | Line 3: | ||
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 <math>999</math>, and the sum of the three numbers formed by reading top to bottom is <math>99</math>. The grid below is an example of such an arrangement because <math>8+991=999</math> and <math>9+9+81=99</math>. | 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 <math>999</math>, and the sum of the three numbers formed by reading top to bottom is <math>99</math>. The grid below is an example of such an arrangement because <math>8+991=999</math> and <math>9+9+81=99</math>. | ||
− | < | + | <cmath> |
\begin{array}{|c|c|c|} \hline | \begin{array}{|c|c|c|} \hline | ||
0 & 0 & 8 \\ \hline | 0 & 0 & 8 \\ \hline | ||
− | 9 & 9 & 1\\ \hline | + | 9 & 9 & 1 \\ \hline |
\end{array} | \end{array} | ||
− | </ | + | </cmath> |
==Solution 1== | ==Solution 1== | ||
Line 30: | Line 30: | ||
</math> | </math> | ||
− | We want <math>10(a+b+c) + (9-a+9-b+9-c) = | + | 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 |
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/nKRfXAHaQvA | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2024|num-b=2|num-a=4|n=II}} | ||
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:08, 5 March 2024
Contents
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 , and the sum of the three numbers formed by reading top to bottom is
. The grid below is an example of such an arrangement because
and
.
Solution 1
Consider this table:
We note that , because
, meaning it never achieves a unit's digit sum of
otherwise. Since no values are carried onto the next digit, this implies
and
. We can then simplify our table into this:
We want , or
, or
. Since zeroes are allowed, we just need to apply stars and bars on
, to get
. ~akliu
Video Solution
~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.