# 2013 AMC 10A Problems/Problem 14

## Problem

A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$. How many edges does the remaining solid have? $\textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad$

## Solution 1

We can use Euler's polyhedron formula that says that $F+V=E+2$. We know that there are originally $6$ faces on the cube, and each corner cube creates $3$ more. $6+8(3) = 30$. In addition, each cube creates $7$ new vertices while taking away the original $8$, yielding $8(7) = 56$ vertices. Thus $E+2=56+30$, so $E=\boxed{\textbf{(D) }84}$

## Solution 2

The removal of each cube adds nine additional edges to the solid. Since a cube initially has $12$ edges and there are eight vertices, the number of edges will be $12 + 9 \times 8 = \boxed{\textbf{(D) } 84}$.

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