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Difference between revisions of "2024 AMC 10A Problems/Problem 1"
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== Solution 4 (Solution 1 but distrubutive) ==
 
== Solution 4 (Solution 1 but distrubutive) ==
 
Note that <math>9901\cdot101=9901\cdot100+9901=9901=990100+9901=1000001</math> and  <math>99\cdot10101=100\cdot10101-10101=1010100-10101=999999</math>, therefore the answer is <math>1000001-999999=\boxed{\text{(A) }2}</math> ~Tacos_are_yummy_1
 
Note that <math>9901\cdot101=9901\cdot100+9901=9901=990100+9901=1000001</math> and  <math>99\cdot10101=100\cdot10101-10101=1010100-10101=999999</math>, therefore the answer is <math>1000001-999999=\boxed{\text{(A) }2}</math> ~Tacos_are_yummy_1
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−
== Video Solution 1 by Power Solve ==
 
−
https://www.youtube.com/watch?v=j-37jvqzhrg
 
  
 
== Video Solution by Daily Dose of Math ==
 
== Video Solution by Daily Dose of Math ==
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~Thesmartgreekmathdude
 
~Thesmartgreekmathdude
 +
 +
== Video Solution 1 by Power Solve ==
 +
https://www.youtube.com/watch?v=j-37jvqzhrg
  
 
==See also==
 
==See also==

Revision as of 21:41, 8 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 (Process of Elimination) (Not recommended)

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 $\boxed{\textbf{(A)}}$ ~mathkiddus

Solution 4 (Solution 1 but distrubutive)

Note that $9901\cdot101=9901\cdot100+9901=9901=990100+9901=1000001$ and $99\cdot10101=100\cdot10101-10101=1010100-10101=999999$, therefore the answer is $1000001-999999=\boxed{\text{(A) }2}$ ~Tacos_are_yummy_1

Video Solution by Daily Dose of Math

https://youtu.be/Z76bafQsqTc

~Thesmartgreekmathdude

Video Solution 1 by Power Solve

https://www.youtube.com/watch?v=j-37jvqzhrg

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

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