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

Revision as of 10:39, 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.

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 answer with $2$ in the units digit is $\textbf{(A)}$ , 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 (⚡️ 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]]}}
+== Problem ==
+What is the value of <math>9901\cdot101-99\cdot10101?</math>
+<math>\textbf{(A)}~2\qquad\textbf{(B)}~20\qquad\textbf{(C)}~200\qquad\textbf{(D)}~202\qquad\textbf{(E)}~2020</math>
+== Solution 1 (Direct Computation) ==
+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 2 (Distributive Property) ==
+We have
+<cmath>\begin{align*}
+\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*}</cmath>
+~MRENTHUSIASM
+== Solution 3 (Solution 1 but Distributive) ==
+Note that <math>9901\cdot101=9901\cdot100+9901=990100+9901=1000001</math> and  <math>99\cdot10101=100\cdot10101-10101=1010100-10101=999999</math>, therefore the answer is <math>1000001-999999=\boxed{\textbf{(A) }2}</math>.
+~Tacos_are_yummy_1
+== Solution 4 (Modular Arithmetic) ==
+Evaluating the given expression <math>\pmod{10}</math> yields <math>1-9\equiv 2 \pmod{10}</math>, so the answer is either <math>\textbf{(A)}</math> or <math>\textbf{(D)}</math>. Evaluating <math>\pmod{101}</math> yields <math>0-99\equiv 2\pmod{101}</math>. Because answer <math>\textbf{(D)}</math> is <math>202=2\cdot 101</math>, that cannot be the answer, so we choose choice <math>\boxed{\textbf{(A) }2}</math>.
+== Solution 5 (Process of Elimination) ==
+We simply look at the units digit of the problem we have (or take mod <math>10</math>)
+<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>.
+~[[User:Mathkiddus|mathkiddus]]
+== 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)==
+Let <math>x=100</math>. Then, we have
+\begin{align*}
+\cdot 9901=(x+1)\cdot (x^2-x+1)=x^3+1, \\
+\cdot 10101=(x-1)\cdot (x^2+x+1)=x^3-1.
+\end{align*}
+Then, the answer can be rewritten as <math>(x^3+1)-(x^3-1)= \boxed{\textbf{(A) }2}.</math>
+~erics118
+==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 ⚡️) ==
+https://youtu.be/RODYXdpipdc
+<i>~Education, the Study of Everything </i>
+== 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
+==See also==
+{{AMC10 box|year=2024|ab=A|before=First Problem|num-a=2}}
+{{AMC12 box|year=2024|ab=A|before=First Problem|num-a=2}}
+{{MAA Notice}}

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