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

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Solution by [[User:Juwushu|juwushu]].
 
Solution by [[User:Juwushu|juwushu]].
  
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The better solution was made by someone else [[User:Juwushu|juwushu]]
  
 
== Solution 2 (Distributive Property) ==
 
== Solution 2 (Distributive Property) ==

Revision as of 16:01, 2 November 2025

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

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 (Direct Computation)

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


A better method, that may be faster actually, would be to multiply 9901 by 100, and then add 9901, then multiply 99 by 10000, 100, and 1, then add them all up. So it would be [($9901\cdot100$) + 9901] -- [($99\cdot10000$) + ($99\cdot100$) + 99] The difference is still $\boxed{\textbf{(A) }2}.$


Solution by juwushu.

The better solution was made by someone else juwushu

Solution 2 (Distributive Property)

We have \begin{align*} 9901\cdot101-99\cdot10101 &= (10000-99)\cdot101-99\cdot(10000+101) \\ &= 10000\cdot101-99\cdot101-99\cdot10000-99\cdot101 \\ &= (10000\cdot101-99\cdot10000)-2\cdot(99\cdot101) \\ &= 2\cdot10000-2\cdot9999 \\ &= \boxed{\textbf{(A) }2}. \end{align*} ~MRENTHUSIASM

Solution 3 (Solution 1 but Distributive)

Note that $9901\cdot101=9901\cdot100+9901=990100+9901=1000001$ and $99\cdot10101=100\cdot10101-10101=1010100-10101=999999$, therefore the answer is $1000001-999999=\boxed{\textbf{(A) }2}$.

~Tacos_are_yummy_1

Solution 4 (Modular Arithmetic)

Evaluating the given expression $\pmod{10}$ yields $1-9\equiv 2 \pmod{10}$, so the answer is either $\textbf{(A)}$ or $\textbf{(D)}$. Evaluating $\pmod{101}$ yields $0-99\equiv 2\pmod{101}$. 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 answers with $2$ in the units digit are $\textbf{(A)}$ and $\textbf{(D)}$, we can then continue if you are desperate to use guess and check or an 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

Solution 9 (Estimation) *🔥VERY FAST🔥*

Notice that the answer choices are significantly different in value. This allows us to estimate the answer. $9901$ is about $10000$, and $101$ is about $100$. $99$ is about $100$, and $10101$ is about $10000$. Computing, we get $10000 \cdot 100-100 \cdot 10000 = 0$. The closest answer to this estimation is $\boxed{\textbf{(A) }2}$.

Solution 10

We can see that the units digit of the expression is $1 - 9 \Rightarrow 11 - 9 \Rightarrow 2$, elimination options B, C, and E. Next, notice that $(9901)(101)$ is divisible by 101 while $(99)(10101)$ is not divisible by 101 (to see this, notice that 101 is prime, and $10101 = 10100 + 1$, so is not divisible by 101). This means that the final answer is not divisible by 101, eliminating $\textbf{(D)}$, so the answer is $\boxed{\textbf{(A) }2}$.

Video Solution(Don't do the actual computation- be fast by taking mods!)

https://youtu.be/l3VrUsZkv8I

~MC

Video Solution by Central Valley Math Circle

https://youtu.be/eLs748wDmMs

~mr_mathman

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

Video Solution by TheBeautyofMath

For AMC 10: https://youtu.be/uKXSZyrIOeU

For AMC 12: https://youtu.be/zaswZfIEibA

~IceMatrix

Video Solution by Dr. David

https://youtu.be/aWu4BJMn9oc

Video Solution by yjtest (2 Solutions, Good Approaches for Competitions)

https://www.youtube.com/watch?v=CSR-edmK52I

Video solution by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=0s

See Also

2024 AMC 10A (Problems • Answer Key • Resources)
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2023 AMC 10B Problems 		Followed by
2024 AMC 10B Problems
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All AMC 10 Problems and Solutions

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