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.

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