481 is a multiple of 37, because $481=37\times 13$. If the digits of 481 are rotated so that the leading digit moves to the back, which of these is the result — and also a multiple of 37?

Moving the 4 from the front to the back turns 481 into 814, and $814=37\times 22$. The property is no accident: $10\times (100a+10b+c)=(100b+10c+a)+999a$, and $999=27\times 37$, so rotating a three-digit multiple of 37 always produces another one. Rotating once more gives $148=37\times 4$. The trap $841=29^2$ merely REVERSES the digits, a different move that destroys the property. And $418=2\times 11\times 19$ and $184=8\times 23$ are rearrangements of the digits that are not rotations at all.