2024 AMC 10A Problems/Problem 1

Revision as of 15:13, 8 November 2024 by Juwushu (talk | contribs) (Solution 1)

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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