Question

If j , k , and n are consecutive integers such that 0 < j < k < n and the units (ones) digit of the product jn is 9, what is the units digit of k ?

Answer 1

The units digit of k is `0`

Explanation:

If j, k, and n are consecutive integers then
`rarr j=k-1` and `n=k+1`

We are told that the units digit of `jn` is 9.

`jn= (k^2-1)`
So
the units digit of `(k^2-1)` is `9`

`rarr` the units digit of `k^2` is `0`

all numbers whose square has a units digit of `0` have a units digit of `0` themselves.