2018 AMC 10B Problems/Problem 1

Revision as of 19:43, 22 December 2020 by Austindragonhood (talk | contribs) (Problem)

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 $\boxed{A}$.

Video Solution

https://youtu.be/o5MUHOmF1zo

~savannahsolver

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2018 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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