1993 AMC 12 Problem 22

Question

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 the numbers assigned to the three blocks on which it rests. Find the smallest possible number which could be assigned to the top block.

55

83

114

137

144

Answer: C

Concepts:weighted sum rearrangement principle triangular stack

Difficulty rating: 1960

Small hint:

Determine how many times each bottom block contributes to the top

Big hint:

The center, six edge positions, and three corner positions have coefficients

6,

3,

and

1

Solution:

Expanding the sums layer by layer, the top value is

6c+3(e_1+e_2+e_3

{}+e_4+e_5+e_6)

{}+(v_1+v_2+v_3),

where

c

is the center bottom block, the

e_i

are the six non-corner boundary blocks, and the

v_i

are the three corner blocks. To minimize this weighted sum, assign

1

to the coefficient-

6

position,

2,

\ldots,

7

to the coefficient-

3

positions, and

8,

9,

10

to the coefficient-

1

positions. The minimum is

6+3(2+3+4

{}+5+6+7)

{}+(8+9+10)=114.

Thus the correct answer is C.

1993 AMC 12 Problem 22

Answer: C

Difficulty rating: 1960

Small hint:

Big hint:

Solution:

Problem 22 in other years