Question

What is the unit digit of 7^349

Answer 1

Let
`u(n)` be the units digits of the number
`n`.

Clearly,
`u(7)=7`.

Since
`7^2=49`, we have
`u(7^2)=9`.

Next,
`7^3=7(49)=7(40+9)`, so only the product
`7*9` will contribute to the value of
`u(7^3)`; indeed,
`7*9=63`, so
`u(7^3)=3`.

Next,
`7^4=7(280+63)=7(340+3)`, and
`7*3=21`, so by the same reasoning as before we have
`u(7^4)=1`.

Continuing in this pattern, we find that


`u(7^5)=7`

`u(7^6)=9`

`u(7^7)=3`

`u(7^8)=1`

`u(7^9)=7`

and so on, with the repeating pattern of
`u(7^n)=\{7,9,3,1,7,9,3,1,\ldots\}` with a period of 4. This means that
`u(7^n)=u(7^m)`, where
`7^n\equiv m\mod4`.

Now,
`349=348+1=4*87+1`, i.e.
`349\equiv1\mod4`, so


`u(7^(349))=u(7^1)=7`