If j, k, and n are consecutive integers such that 0 less than j less than k less than n and the units (ones) digit of the product j times n is 9, what is the units digit of k

Question

anonymous · Answer

$j,k,n$ are consecutive integers, so if $j$ is the smallest, then $k=j+1$ and $n=j+2$.

The product $jn$ has a 9 in the ones digit, which means the ones digits of these two numbers are factors of 9, either 3 and 3, or 9 and 1.

Suppose the ones digit of both $n$ and $j$ is 3. To simplify, suppose $j=3$. But then $n=j+2=5$, which contradicts our assumption.

This means the ones digit of $j$ is either 1 or 9, and for $n$ it's either 9 or 1. Clearly, we have the first case. For instance, if $j=9$, then $n=j+2=11$ (note the ones digits).

This means $k=j+1=10$, and so the ones digit of $k$ is 0.