2024 AMC 10A Problems/Problem 1

The following problem is from both the 2024 AMC 10A #1 and 2024 AMC 12A #1, so both problems redirect to this page.

1 Solution 1 (Distributive Property but faster)
2 Solution 2 (Distributive Property)
3 Solution 3 (Solution 1 but Distributive)
4 Solution 4 (Modular Arithmetic)
5 Solution 5 (Process of Elimination)
6 Solution 6 (Faster Distribution)
7 Solution 7 (Cubes)
8 Solution 8 (Super Fast)
9 Video Solution (⚡️ 1 min solve ⚡️)
10 Video Solution by Pi Academy
11 Video Solution by FrankTutor
12 Video Solution Daily Dose of Math
13 Video Solution 1 by Power Solve
14 Video Solution by SpreadTheMathLove
15 Video Solution by Math from my desk

Solution 1 (Distributive Property but faster)

$\begin{align*} \\ 9901 \cdot 101 - 99 \cdot 10101 &= (9900 + 1) \cdot 101 - 99 \cdot (10100 +1) \\ &= 999900 + 101 - 999900 - 99 \\ &= 101 - 99 = \boxed{\textbf{(A) }2} \end{align*}$ ~sorebrek

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 $101-99=\boxed{\textbf{(A) }2}$ .