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

Revision as of 02:07, 31 January 2025

The following problem is from both the 2024 AMC 10A #1 and 2024 AMC 12A #1, so both problems redirect to this page.

~Education, the Study of Everything

~Thesmartgreekmathdude

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 {{duplicate|[[2024 AMC 10A Problems/Problem 1|2024 AMC 10A #1]] and [[2024 AMC 12A Problems/Problem 1|2024 AMC 12A #1]]}}
-==Solution 8 (Super Fast)==
-It's not hard to observe and express <math>9901</math> into <math>99\cdot100+1</math>, and <math>10101</math> into <math>101\cdot100+1</math>.
-We then simplify the original expression into <math>(99\cdot100+1)\cdot101-99\cdot(101\cdot100+1)</math>, which could then be simplified into <math>99\cdot100\cdot101+101-99\cdot100\cdot101-99</math>, which we can get the answer of <math>101-99=\boxed{\textbf{(A) }2}</math>.
-~RULE101
 == Video Solution (⚡️ 1 min solve ⚡️) ==