# 1993 AHSME Problems/Problem 22

## Problem

Twenty cubical blocks are arranged as shown. First, 10 are arranged in a triangular pattern; then a layer of 6, arranged in a triangular pattern, is centered on the 10; then a layer of 3, arranged in a triangular pattern, is centered on the 6; and finally one block is centered on top of the third layer. The blocks in the bottom layer are numbered 1 through 10 in some order. Each block in layers 2,3 and 4 is assigned the number which is the sum of numbers assigned to the three blocks on which it rests. Find the smallest possible number which could be assigned to the top block. $\text{(A) } 55\quad \text{(B) } 83\quad \text{(C) } 114\quad \text{(D) } 137\quad \text{(E) } 144$

## Solution

The value assigned at the top is the weighted sum of the values in the cubes of a given level. The weight for a given cube is the sum of the weights of the blocks which touch it above, since the number in a cube will get incorporated into the overall sum via each of those blocks, according to the weight of each block. Thus, if we think about how the weights propagate down the block pyramid, its sort of like a three dimensional Pascal's triangle.

The first layer is $1$

The second layer is $| \, \, 1 \\ |1 \, 1$

The third layer is $| \,\,\,\, 1 \\| \,\, 2 \, 2 \\| 1 \,2 \, 1$

The fourth layer is $| \,\,\,\,\,\, 1 \\| \,\,\,\, 3 \, 3 \\| \,\, 3 \,6 \, 3 \\| 1\, 3\, 3\, 1$

The top level sum is minimized if we associate the block with weight 6 with the smallest number 1, the blocks with weight 1 with the largest numbers 8, 9 and 10, and the rest of the blocks with weight 3 with the rest of the numbers. The sum is $6\cdot 1 + 3 \cdot (2+3+4+5+6+7 ) + 1\cdot (8+9+10) = 114$ so the answer is $\fbox{C}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 