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

Revision as of 20:17, 9 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 $9901\cdot101=9901\cdot100+9901=9901=990100+9901=1000001$ 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, fast)

Evaluating the given expression $\pmod{10}$ yields $1-9\equiv 2 \pmod{10}$ , so the answer is either (A) or (D). Evaluating $\pmod{101}$ yields $0-99\equiv 2\pmod{101}$ . Because answer (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 $\textbf{(A)}$ 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 $\boxed{\textbf{(A) }2}$ . ~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

Video Solution by Pi Academy

https://youtu.be/GPoTfGAf8bc?si=JYDhLVzfHUbXa3DW

Video Solution

https://youtu.be/Z76bafQsqTc

~Thesmartgreekmathdude

Video Solution 1 by Power Solve

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

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

@@ Line 31: / Line 31: @@
 Evaluating the given expression <math>\pmod{10}</math> yields <math>1-9\equiv 2 \pmod{10}</math>, so the answer is either (A) or (D). Evaluating <math>\pmod{101}</math> yields <math>0-99\equiv 2\pmod{101}</math>. Because answer (D) 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) (Not recommended) ==
+== 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 2 in the units digit is <math>\textbf{(A)}</math> or <math>\textbf{(D)}</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]]
-~mathkiddus
 == Solution 6 (Faster Distribution) ==

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