# Difference between revisions of "2018 AMC 10B Problems/Problem 1"

Problem: Suppose we write down the smallest (positive) $2$-digit, $3$-digit, and $4$-digit multiples of $8$.

## Solution 1

The area of the pan is $20\cdot18$ = $360$. Since the area of each piece is $4$, there are $\frac{360}{4} = 90$ pieces. Thus, the answer is $\boxed{A}$.

## Solution 2

By dividing each of the dimensions by $2$, we get a $10\times9$ grid which makes $90$ pieces. Thus, the answer is $\boxed{A}$.

~savannahsolver