A certain number ends with the digit 2
. Moving the 2
from the end of the number to its front doubles it. Can you find this number?
(Hint: it's quite large)
I think the easiest way to get the solution is to realize that if 2 is in the ones place before doubling, then a 4 must be in the ones place after doubling. Then it becomes simple addition.
.... X 4 2
... X 4 2 +
.... Y X 4
The first solution is when X finally becomes a 1 without a carryover. The number is 105263157894736842. Unless I made a sloppy addition error. This solution is not unique and there is in fact infinitely many of them.
The second solution is continuing the process again until the number 1 is reached once more without a carryover. This number is 105263157894736842105263157894736842, (unless I have made an addition error in the first part). This number has all the same properties of the first. The process can, without a doubt, go on as long as you have room to write all of the digits down.