Question

Say N is a natural number and M is it's reverse.
Say M=Nk, where k is a natural number.
Prove that whenever such an equality holds, k can only be 1,4 or 9(a perfect square).
Example : 2178*4=8712
or 1089*9=9801

Answer 1

Correction: M contains the same number of digits as N.

Answer 2

Slight fix, \(M \ge N\) and \(M=kN\)

\[M \equiv N \mod 3\]\[M \equiv N \mod 9\]

It appears that \(k=9\ \implies M \equiv 0 \mod 3\) and \(M \equiv 0 \mod 9\)

Answer 3

I remember thinking long and hard about this problem, maybe there are some thoughts in here that might help: http://math.stackexchange.com/questions/754211/is-there-a-power-of-2-that-written-backward-is-a-power-of-5

Answer 4

.

Answer 5

I guess "M contains the same number of digits as N." means the same thing as saying that N isn't divisible by 10.

Answer 6

yus.

Answer 7

I think the trick here is to show M/N is a square number,
since the number of digits are same, it is by default 1,4 or9

Answer 8

That is really clever !