Difference between revisions of "2024 AIME II Problems/Problem 3"
Iwowowl253 (talk | contribs) m (→Solution 1) |
Iwowowl253 (talk | contribs) (→Solution 1) |
||
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 | + | 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 |
==Video Solution== | ==Video Solution== |
Revision as of 17:17, 14 February 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 |
[[Category:]]
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.