Question

A triangular array of squares has one square in the first row, two in the second, and, in general, $k$ squares in the $k$ th row for $1 \le k \le 11.$ With the exception of the bottom row, each square rests on two squares in the row immediately below, as illustrated in the figure. In each square of the eleventh row, a $0$ or a $1$ is placed. Numbers are then placed into the other squares, with the entry for each square being the sum of the entries in the two squares below it. For how many initial distributions of $0$ 's and $1$ 's in the bottom row is the number in the top square a multiple of $3?$

Answer

Solution:

Label the bottom-row entries $x_0, x_1, \ldots, x_{10}.$ Since each square is the sum of the two below it, the contributions accumulate with Pascal's-triangle weights: the top square equals $\sum_{i=0}^{10} \binom{10}{i} x_i.$

Modulo $3,$ direct checking (or Lucas' theorem with $10 = 101_3$ ) shows $\binom{10}{i} \equiv 0$ for $2 \le i \le 8,$ while $\binom{10}{0} = \binom{10}{10} = 1$ and $\binom{10}{1} = \binom{10}{9} = 10 \equiv 1.$ So the top square is a multiple of $3$ exactly when $x_0 + x_1 + x_9 + x_{10} \equiv 0 \pmod 3.$

For $0$ / $1$ entries this sum is $0$ or $3:$ either all four are $0$ (one way) or exactly three are $1$ (four ways), for $5$ choices. The remaining seven entries $x_2, \ldots, x_8$ are free, so the count is $5 \cdot 2^7 = 640.$

2007 AIME II Problem 13

Solution:

Problem 13 in Other Years