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

(Problem)
(Solution 1)
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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>.
 
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 ==
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Problem:
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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>.
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What is the sum of these three numbers?
  
 
== Solution 2 ==
 
== Solution 2 ==

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 $8$.

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?

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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