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 21:01, 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
.