2024 AMC 10A Problem 1

Long multiplication would work, but it is slow and easy to get wrong. The numbers $101$ and $99$ are both right next to $100$, and multiplying by $100$ is easy (just add two zeros). So use that.

1. Write $101 = 100 + 1$. Then $9901 \cdot 101$ is "$9901$ times $100$, plus one more $9901$":
$$9901 \cdot 101 = 990100 + 9901 = 1000001.$$
2. Write $99 = 100 - 1$. Then $99 \cdot 10101$ is "$10101$ times $100$, minus one $10101$":
$$99 \cdot 10101 = 1010100 - 10101 = 999999.$$
3. Subtract: $1000001 - 999999 = 2$.

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

Multiplying by a number close to $100$ (or $1000$, or $10$) is the same as multiplying by $100$ and then fixing up the small difference. That turns a hard multiplication into one easy addition or subtraction. Whenever you see factors like $99$, $101$, $999$ or $1001$, try this first.

You can rule out most choices without finishing the arithmetic. Look only at the last digits: $9901 \cdot 101$ ends in $1 \cdot 1 = 1$, and $99 \cdot 10101$ ends in $9 \cdot 1 = 9$. A number ending in $1$ minus a number ending in $9$ ends in $2$ (think $11 - 9 = 2$). So the answer is (A) $2$ or (D) $202$.

Now estimate: $9901 \cdot 101$ is about $10000 \cdot 100 = 1{,}000{,}000$, and $99 \cdot 10101$ is also about $100 \cdot 10000 = 1{,}000{,}000$. The two products are very close, so the small answer $2$ is the natural one, and the exact work above confirms it.