2024 AMC 10A Problems/Problem 1: Difference between revisions

Latest revision as of 17:02, 6 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 $\boxed{\textbf{(A) }2}.$

Solution by 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

@@ Line 3: / Line 3: @@
 == Problem ==
-. If ojas had two chodes and ate preetam's chode how many chodes would be left?
+What is the value of <math>9901\cdot101-99\cdot10101?</math>
-<img src="pic_trulli.jpg" alt="Italian Trulli">
+<math>\textbf{(A)}~2\qquad\textbf{(B)}~20\qquad\textbf{(C)}~200\qquad\textbf{(D)}~202\qquad\textbf{(E)}~2020</math>
-== Solution 0 (Gooning) ==
+== Solution 1 (Direct Computation) ==
-With skibidi toilet and Shaesty window incident you can solve the problem. <math>9901\cdot101</math> evaluates to <math>1000001</math> and <math>99\cdot10101</math> evaluates to <math>999999</math>. The difference is <math>\boxed{\textbf{(A) }2}.</math>
+The likely fastest method will be direct computation. <math>9901\cdot101</math> evaluates to <math>1000001</math> and <math>99\cdot10101</math> evaluates to <math>999999</math>. The difference is <math>\boxed{\textbf{(A) }2}.</math>
 Solution by [[User:Juwushu|juwushu]].
-== Solution 1 (Distributive Property but faster) ==
-<cmath>\begin{align*} \\
-\cdot 101 - 99 \cdot 10101 &= (9900 + 1) \cdot 101 - 99 \cdot (10100 +1) \\
-&= 999900 + 101 - 999900 - 99 \\
-&= 101 - 99 = \boxed{\textbf{(A) }2}
-\end{align*}</cmath>
-~sorebrek
 == Solution 2 (Distributive Property) ==
@@ Line 41: / Line 33: @@
 == Solution 5 (Process of Elimination) ==
-We simply look at the units digit of the problem we have (or take mod <math>10</math>)
+We simply look at the units digit of the problem we have (or take mod <imath>10</imath>)
 <cmath>9901\cdot101-99\cdot10101 \equiv 1\cdot1 - 9\cdot1 = 2 \mod{10}.</cmath>
-Since the only answer with <math>2</math> in the units digit is <math>\textbf{(A)}</math>, We can then continue if you are desperate to use guess and check or a actually valid method to find the answer is <math>\boxed{\textbf{(A) }2}</math>.
+Since the only answers with <imath>2</imath> in the units digit are <imath>\textbf{(A)}</imath> and <imath>\textbf{(D)}</imath>, we can then continue if you are desperate to use guess and check or an actually valid method to find the answer is <imath>\boxed{\textbf{(A) }2}</imath>.
 ~[[User:Mathkiddus|mathkiddus]]
@@ Line 76: / Line 68: @@
 ~RULE101
+==Solution 9 (Estimation) *🔥VERY FAST🔥*==
+Notice that the answer choices are significantly different in value. This allows us to estimate the answer. <math>9901</math> is about <math>10000</math>, and <math>101</math> is about <math>100</math>. <math>99</math> is about <math>100</math>, and <math>10101</math> is about <math>10000</math>. Computing, we get <math>10000 \cdot 100-100 \cdot 10000 = 0</math>. The closest answer to this estimation is <math>\boxed{\textbf{(A) }2}</math>.
+==Solution 10==
+We can see that the units digit of the expression is <math>1 - 9 \Rightarrow 11 - 9 \Rightarrow 2</math>, elimination options B, C, and E. Next, notice that <math>(9901)(101)</math> is divisible by 101 while <math>(99)(10101)</math> is not divisible by 101 (to see this, notice that 101 is prime, and <math>10101 = 10100 + 1</math>, so is not divisible by 101). This means that the final answer is not divisible by 101, eliminating <math>\textbf{(D)}</math>, so the answer is <math>\boxed{\textbf{(A) }2}</math>.
+==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 ⚡️) ==
@@ Line 103: / Line 112: @@
 ==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==
+{{AMC10 box|year=2024|ab=A|before=[[2023 AMC 10B Problems]]|after=[[2024 AMC 10B Problems]]}}
+* [[AMC 10]]
+* [[AMC 10 Problems and Solutions]]
+* [[Mathematics competitions]]
+* [[Mathematics competition resources]]
+{{MAA Notice}}

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