Difference between revisions of "2024 AIME II Problems/Problem 3"
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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 | 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 | ||
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| + | ==See also== | ||
| + | {{AIME box|year=2024|num-b=2|num-a=4|n=II}} | ||
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| + | [[Category:]] | ||
| + | {{MAA Notice}} | ||
Revision as of 20:53, 8 February 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
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. 
