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Difference between revisions of "2024 AMC 10A Problems/Problem 1"
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{{duplicate|[[2024 AMC 10A Problems/Problem 1|2024 AMC 10A #1]] and [[2024 AMC 12A Problems/Problem 1|2024 AMC 12A #1]]}}
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== Problem ==
 
== Problem ==
  
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What is the value of <math>9901\cdot101-99\cdot10101?</math>
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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>
 
<math>\textbf{(A)}~2\qquad\textbf{(B)}~20\qquad\textbf{(C)}~200\qquad\textbf{(D)}~202\qquad\textbf{(E)}~2020</math>
  
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== Solution 1 ==
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== Solution 1 (Direct Computation) ==
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The likely fastest method will be straight computation. <math>9901\cdot101</math> evaluates to <math>1000001</math> and <math>99\cdot10101</math> evaluates to <math>999999</math>. The difference is \boxed{\textbf{(A) }2}
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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]].
 +
 
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== Solution 2 (Distributive Property) ==
 +
We have
 +
<cmath>\begin{align*}
 +
9901\cdot101-99\cdot10101 &= (10000-99)\cdot101-99\cdot(10000+101) \\
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&= 10000\cdot101-99\cdot101-99\cdot10000-99\cdot101 \\
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&= (10000\cdot101-99\cdot10000)-2\cdot(99\cdot101) \\
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&= 2\cdot10000-2\cdot9999 \\
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&= \boxed{\textbf{(A) }2}.
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\end{align*}</cmath>
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~MRENTHUSIASM
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== Solution 3 (Solution 1 but Distributive) ==
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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
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== 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]]
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== Solution 6 (Faster Distribution) ==
 +
Observe that <math>9901=9900+1=99\cdot100+1</math> and <math>10101=10100+1=101\cdot100+1</math>
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<cmath>\begin{align*}
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\Rightarrow9901\cdot101-99\cdot10101 & = ((9900\cdot101)+(1\cdot101))-((99\cdot10100)+(99\cdot1)) \\
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&=(99\cdot100\cdot101)+101-(99\cdot100\cdot101)-99 \\
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&=101-99 \\
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&=\boxed{\textbf{(A) }2}.
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\end{align*}</cmath>
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 +
~laythe_enjoyer211
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==Solution 7 (Cubes)==
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Let <math>x=100</math>. Then, we have
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\begin{align*}
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101\cdot 9901=(x+1)\cdot (x^2-x+1)=x^3+1, \\
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99\cdot 10101=(x-1)\cdot (x^2+x+1)=x^3-1.
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\end{align*}
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Then, the answer can be rewritten as <math>(x^3+1)-(x^3-1)= \boxed{\textbf{(A) }2}.</math>
 +
 
 +
~erics118
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==Solution 8 (Super Fast)==
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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
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 +
== 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
 +
 
 +
==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}}

Latest revision as of 11:26, 27 November 2024

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 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

See also

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. AMC logo.png

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