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AHSME 1993 Problem 22

    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. (A) 55(B) 83(C) 114(D) 137(E) 144\text{(A) } 55 \text{(B) } 83 \text{(C) } 114 \text{(D) } 137 \text{(E) } 144

    Dataset text NuminaMath-1.5. Quoted with source.

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    114

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