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Problem 554

AMC 10/12, early questions

Number theory Difficulty 3.5 Find the answer CEMC Fermat · Canada · 2021

If $n$ is a positive integer, the notation $n!$ (read “ $n$ factorial”) is used to represent the product of the integers from 1 to $n$ . That is, $n! = n(n-1)(n - 2) \cdots (3)(2)(1)$ . For example, $4! = 4(3)(2)(1) = 24$ and $1!=1$ . If $a$ and $b$ are positive integers with $b>a$ , the ones (units) digit of $b!-a!$ cannot be

$1$
$3$
$5$
$7$
$9$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

The first few values of $n!$ are $\begin{aligned} 1! & = 1 \\ 2! & = 2(1) = 2\\ 3! & = 3(2)(1) = 6 \\ 4! & = 4(3)(2)(1) = 24 \\ 5! & = 5(4)(3)(2)(1) = 120\end{aligned}$ We note that $\begin{aligned} 2! - 1! & = 1 \\ 4! - 1! & = 23 \\ 3! - 1! & = 5 \\ 5! - 1! & = 119\end{aligned}$ This means that if $a$ and $b$ are positive integers with $b>a$ , then 1, 3, 5, 9 are all possible ones (units) digits of $b!-a!$ .

This means that the only possible answer is choice (D), or 7.

To be complete, we explain why 7 cannot be the ones (units) digit of $b!-a!$ .

For $b!-a!$ to be odd, one of $b!$ and $a!$ is even and one of them is odd.

The only odd factorial is $1!$ , since every other factorial has a factor of 2.

Since $b > a$ , then if one of $a$ and $b$ is 1, we must have $a=1$ .

For the ones (units) digit of $b! - 1$ to be 7, the ones (units) digit of $b!$ must be 8.

This is impossible as the first few factorials are shown above and every greater factorial has a ones (units) digit of 0, because it is a multiple of both 2 and 5.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.