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

Revision as of 18:45, 22 December 2020

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

What is the sum of these three numbers?

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

2018 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2018 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

@@ Line 2: / Line 2: @@
 Suppose we write down the smallest (positive) <math>2</math>-digit, <math>3</math>-digit, and <math>4</math>-digit multiples of <math>8</math>.
-== Solution 1 ==
+Problem:
+Suppose we write down the smallest (positive) <math>2</math>-digit, <math>3</math>-digit, and <math>4</math>-digit multiples of <math>8</math>.
-The area of the pan is <math>20\cdot18</math> = <math>360</math>. Since the area of each piece is <math>4</math>, there are <math>\frac{360}{4} = 90</math> pieces. Thus, the answer is <math>\boxed{A}</math>.
+What is the sum of these three numbers?
 == Solution 2 ==