shubhamsrg
  • shubhamsrg
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
Mathematics
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SOLVED
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katieb
  • katieb
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shubhamsrg
  • shubhamsrg
Correction: M contains the same number of digits as N.
Kainui
  • Kainui
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 \(M \equiv 0 \mod 9\)
Kainui
  • Kainui
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

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ganeshie8
  • ganeshie8
.
Kainui
  • Kainui
I guess "M contains the same number of digits as N." means the same thing as saying that N isn't divisible by 10.
shubhamsrg
  • shubhamsrg
yus.
baru
  • baru
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
ganeshie8
  • ganeshie8
That is really clever !

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