Let the original six digit number be `1abcde,` where `a,b,c,d,e` are the unknown digits. Moving the `1` to the end results in the new six digit number `abcde1.` It is given that

`1abcde`

`xx.....3`

`-----`

`abcde1`

` `

` `

` `

` `

` `

` ` The dots are just there for spacing, so don't let them confuse you; I couldn't figure out how to get the 3 on the right without them.

Anyway, we know that `3e` ends in 1, so `e` must be 7 and we can make the replacement to get

`1abcd7`

`xx.....3`

`-----`

`abcd71`

` `

Then, since we carry the 2 from `7xx3=21,` we know that `3d+2` ends in `7,` and the only `d` that works is `5,` so

`1abc57`

`xx.....3`

`-----`

`abc571`

Finally, we must have `3c+1` ends in `5` (remember the `+1` comes from carrying the one from `5xx3+2=17` ). The only `c` that works is `8,` and the last three digits of the original number are 857.

You can continue this process to get the whole number if you wish.