Problem:
Find the number of ways to place a digit in each cell of a 2×3 grid so that the sum of the two numbers formed by reading left to right is 999, and the sum of the three numbers formed by reading top to bottom is 99. The grid below is an example of such an arrangement because 8+991=999 and 9+9+81=99.
090981
Solution:
Label the cells' digits as shown below.
ABCDEF
Then the two given conditions imply that
A C E+B D F=999andA B+C D+E F=99,
which are equivalent to
(100A+10C+E)+(100B+10D+F)=999and(10A+B)+(10C+D)+(10E+F)=99.
Because there is no carry in the first summation, it follows that A+B=C+D=E+F=9, so
(A+C+E)+(B+D+F)=27.
The second given condition implies that
10(A+C+E)+(B+D+F)=99,
from which A+C+E=8 and B+D+F=19. Observe that once A,C, and E are chosen subject to A+C+E=8, then B,D, and F are fixed and satisfy the given relations. By the sticks-and-stones method, the number of nonnegative integer solutions to A+C+E=8 is (3−18+3−1)=45, which is the requested number of arrangements.
The problems on this page are the property of the MAA's American Mathematics Competitions