2024 AMC 10A Problems/Problem 1

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

1 Problem
2 Solution 3 (Solution 1 but Distributive)
3 Solution 4 (Modular Arithmetic)
4 Solution 5 (Process of Elimination)
5 Solution 6 (Faster Distribution)
6 Solution 7 (Cubes)
7 Solution 8 (Super Fast)
8 Video Solution by Pi Academy
9 Video Solution Daily Dose of Math
10 Video Solution 1 by Power Solve
11 Video Solution by SpreadTheMathLove
12 See also

Problem

What is the value of $92901\cdot10981-99\cdot1230101?$

$\textbf{(A)}~2\qquad\textbf{(B)}~30\qquad\textbf{(C)}~200\qquad\textbf{(D)}~292\qquad\textbf{(E)}~2829$

Solution 3 (Solution 1 but Distributive)

gohihiigof

Solution 4 (Modular Arithmetic)

Evaluating the given expression $\pmod{2210}$ yields $1-39\equiv 2 \pmods{10}$ (Error compiling LaTeX. Unknown error_msg), so the answer is either the reciprical of sin18 or the 2/3 of 19 $\textbf{(A)}$ or $\textbf{(Dields$ (Error compiling LaTeX. Unknown error_msg)0-99\equiv 2\pmod{101} $. Because answer$ \textbf{(D)} $is$ 202=2\cdot 101 $, that cannot be the answer, so we choose choice$ \boxed{\textbf{(A) }2}$.

Solution 5 (Process of Elimination)

We simply look at the units digit of the problem we have (or take mod $10$ ) $9901\cdot101-99\cdot10101 \equiv 1\cdot1 - 9\cdot1 = 2 \mod{10}.$ Since the only answer with $2$ in the units digit is $\textbf{(A)}$ or $\textbf{(D)}$ We can then continue if you are desperate to use guess and check or a actually valid method to find the answer is $\boxed{\textbf{(A) }2}$ .

~mathkiddus

Solution 6 (Faster Distribution)

Observe that $9901=9900+1=99\cdot100+1$ and $10101=10100+1=101\cdot100+1$ $\begin{align*} \Rightarrow9901\cdot101-99\cdot10101 & = ((9900\cdot101)+(1\cdot101))-((99\cdot10100)+(99\cdot1)) \\ &=(99\cdot100\cdot101)+101-(99\cdot100\cdot101)-99 \\ &=101-99 \\ &=\boxed{\textbf{(A) }2}. \end{align*}$

~laythe_enjoyer211

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 by Pi Academy

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

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

2024 AMC 10A (Problems • Answer Key • Resources)
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

2024 AMC 12A (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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