2024 AIME II Problems/Problem 3: Difference between revisions
Iwowowl253 (talk | contribs) |
R00tsofunity (talk | contribs) No edit summary |
||
| Line 41: | 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 17:08, 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
, 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing