2024 AMC 10A Problem 1

Both products are easy once you write $101 = 100 + 1$ and $99 = 100 - 1$.

$$9901\cdot 101 = 9901\cdot 100 + 9901 = 990100 + 9901 = 1000001.$$
$$99\cdot 10101 = 10101\cdot 100 - 10101 = 1010100 - 10101 = 999999.$$

Subtracting, $1000001 - 999999 = 2$.

The answer is $\boxed{\textbf{(A)}\ 2}$.

Numbers like $101$, $99$, $9901$, $10101$ are powers of ten plus or minus something small, so the distributive property turns each product into an addition or subtraction of two numbers you can write down instantly. Whenever a factor is within a few units of a power of ten, split it that way instead of doing long multiplication.

Let $x = 100$. Then $9901 = x^2 - x + 1$ and $101 = x + 1$, so the first product is $x^3 + 1$; likewise $99 = x - 1$ and $10101 = x^2 + x + 1$ give $x^3 - 1$. The difference is $2$. A quick units-digit check ($1\cdot 1 - 9\cdot 1$ ends in $2$) also narrows the choices to (A) or (D).

• Multiplying longhand and misplacing a zero; the two products differ by only 2, so any slip in a digit lands on a wrong choice.
• Trusting the units-digit check alone, which leaves both (A) and (D) possible.