# Difference between revisions of "1994 AJHSME Problems/Problem 6"

## Problem

The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is

$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

## Solution

Within six consecutive integers, there must be a number with a factor of $5$ and an even integer with a factor of $2$. Multiplied together, these would produce a number that is a multiple of $10$ and has a units digit of $\boxed{\text{(A)}\ 0}$.

## Solution 2

We can easily compute the product of the first 6 positive integers: $(1*2*3*4*5*6)=6!=720$ Therefore the units digit must be $\boxed{\text{(A)}\ 0}$.