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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Revision as of 19:59, 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
