Difference between revisions of "2024 AIME II Problems/Problem 3"
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==Solution 2== | ==Solution 2== | ||
| + | Let this be the table: | ||
| + | <math> | ||
| + | \begin{array}{|c|c|c|} \hline | ||
| + | a & b & c \\ \hline | ||
| + | d & e & f\\ \hline | ||
| + | \end{array} | ||
| + | </math> | ||
| + | Then, <math>100(a+d)+10(b+e)+(c+f)=999</math> and <math>10(a+b+c)+(d+e+f)=99</math>. We look at the first equation. Since <math>c</math> and <math>f</math> are digits, <math>c<9</math> and <math>f<9</math>, so <math>c+f<18</math>, which means <math>c+f=9</math> for the last digit to be <math>9</math>. Similarly, <math>b+e=9</math> and <math>a+d=9</math>. | ||
Revision as of 20:01, 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
Solution 2
Let this be the table:
Then, 
and 
. We look at the first equation. Since 
and 
are digits, 
and 
, so 
, which means 
for the last digit to be . Similarly, and .
