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.

Solution 7 (Cubes)

Let $x=100$ . Then, we have \begin{align*} 101\cdot 9901=(x+1)\cdot (x^2-x+1)=x^3+1, \\ 99\cdot 10101=(x-1)\cdot (x^2+x+1)=x^3-1. \end{align*} Then, the answer can be rewritten as $(x^3+1)-(x^3-1)= \boxed{\textbf{(A) }2}.$

~erics118

Solution 8 (Super Fast)

It's not hard to observe and express $9901$ into $99\cdot100+1$ , and $10101$ into $101\cdot100+1$ .

We then simplify the original expression into $(99\cdot100+1)\cdot101-99\cdot(101\cdot100+1)$ , which could then be simplified into $99\cdot100\cdot101+101-99\cdot100\cdot101-99$ , which we can get the answer of $101-99=\boxed{\textbf{(A) }2}$ .

~RULE101

Video Solution (⚡️ 1 min solve ⚡️)

https://youtu.be/RODYXdpipdc

~Education, the Study of Everything

Video Solution by Pi Academy

https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW

Video Solution by FrankTutor

https://www.youtube.com/watch?v=ez095SvW5xI

Video Solution Daily Dose of Math

https://youtu.be/Z76bafQsqTc

~Thesmartgreekmathdude

Video Solution 1 by Power Solve

https://www.youtube.com/watch?v=j-37jvqzhrg

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

Video Solution by Math from my desk

https://www.youtube.com/watch?v=n_G6wi1ulzY

@@ Line 1: / Line 1: @@
 {{duplicate|[[2024 AMC 10A Problems/Problem 1|2024 AMC 10A #1]] and [[2024 AMC 12A Problems/Problem 1|2024 AMC 12A #1]]}}
-== Solution 6 (Faster Distribution) ==
-Observe that <math>9901=9900+1=99\cdot100+1</math> and <math>10101=10100+1=101\cdot100+1</math>
-<cmath>\begin{align*}
-\Rightarrow9901\cdot101-99\cdot10101 & = ((9900\cdot101)+(1\cdot101))-((99\cdot10100)+(99\cdot1)) \\
-&=(99\cdot100\cdot101)+101-(99\cdot100\cdot101)-99 \\
-&=101-99 \\
-&=\boxed{\textbf{(A) }2}.
-\end{align*}</cmath>
-~laythe_enjoyer211
 ==Solution 7 (Cubes)==

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