Difference between revisions of "2024 AMC 10A Problems/Problem 1"

(Solution 1)
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== Problem ==
 
== Problem ==
  
What is the value of <math>9901\cdot101-99\cdot10101?</math>
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What is the value of <cmath>9901\cdot101-99\cdot10101?</cmath>
  
 
<math>\textbf{(A)}~2\qquad\textbf{(B)}~20\qquad\textbf{(C)}~200\qquad\textbf{(D)}~202\qquad\textbf{(E)}~2020</math>
 
<math>\textbf{(A)}~2\qquad\textbf{(B)}~20\qquad\textbf{(C)}~200\qquad\textbf{(D)}~202\qquad\textbf{(E)}~2020</math>
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== Solution 1 ==
 
== Solution 1 ==
 
The likely fastest method will be straight computation. <math>9901\cdot101</math> evaluates to <math>1000001</math> and <math>99\cdot10101</math> evaluates to <math>999999</math>. The difference is <math>\boxed{\textbf{(A) }2}.</math>
 
The likely fastest method will be straight computation. <math>9901\cdot101</math> evaluates to <math>1000001</math> and <math>99\cdot10101</math> evaluates to <math>999999</math>. The difference is <math>\boxed{\textbf{(A) }2}.</math>
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Solution by [[User:Juwushu|juwushu]].
 
Solution by [[User:Juwushu|juwushu]].
  

Revision as of 15:15, 8 November 2024

Problem

What is the value of \[9901\cdot101-99\cdot10101?\]

$\textbf{(A)}~2\qquad\textbf{(B)}~20\qquad\textbf{(C)}~200\qquad\textbf{(D)}~202\qquad\textbf{(E)}~2020$

Solution 1

The likely fastest method will be straight computation. $9901\cdot101$ evaluates to $1000001$ and $99\cdot10101$ evaluates to $999999$. The difference is $\boxed{\textbf{(A) }2}.$

Solution by juwushu.

See also

2024 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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All AMC 10 Problems and Solutions

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