A number ending in four becomes four times larger when the end four is removed and placed at the front?

This can be solved by expressing parts of the number separately.

You already know that the last digit of your starting number is 4. You also know that if you take that last digit, and move it to the front of the number, the result is four times the original number. What we don't know is how many digits the rest of the number is, or what it's value is. For now, let's call those remaining digits "A".

In that case, we can say that our original number is 10A + 4, and our resulting number is 4 × 10x + A where x is the number of digits used in the number A.

We also know that our resulting number is four times the value of the first one, and with that we can assemble the equation:

4(10A + 4) = 4 × 10x + A

and simplify it:

∴40A + 16 = 4 × 10x + A

∴39A = 4 × 10x - 16

∴A = (4 × 10x - 16) / 39

All we have to do now is try different values of x until we find one that results in A being an integer value.

Let's see if we can do that with a two digit number:

A = (4 × 102 - 16) / 39

∴ A = (400 - 16) / 39

∴ A = 384 / 39

Of course that doesn't actually work, because it gives us a fraction. We may have to go through several numbers before we find one that works, but that can be done pretty quickly. We just need to add a zero to our multiple of 4, and keep testing until we find one that gives us an integer:

(4000 - 16) / 39 = 3984 / 39 ≈ 102.15

(40000 - 16) / 39 = 39984 / 39 ≈ 1025.23

(400000 - 16) / 39 = 399984 / 39 = 10256

And there we have the number we want. A is equal to 10256, and we can now say that the original number we're looking for is:

102564

And if we test it to be sure:

102564 × 4 = 410256

We can see that this does indeed work.