I saw this puzzle on this address.






I know that first $4$ digits are $4864$. How could I find last $2$ digits?


1 Answer 1


Surprisingly, the last $2$ digits will be

not $2$ digits, but $3$ digits! Which is $821$.


For $A + B + C$, the last $2$ (err.. $3$) digits will be the reverse of $A^2 \times (C-B)$.

$3+5+7$ will be $rev(3^2 \times (7-5)) = rev(18) = 81$.
$4+5+6$ will be $rev(4^2 \times (6-5)) = rev(16) = 61$.
$3+4+7$ will be $rev(3^2 \times (7-4)) = rev(27) = 72$.
$9+4+5$ will be $rev(9^2 \times (5-4)) = rev(81) = 18$.
$8+6+8$ will be $rev(8^2 \times (8-6)) = rev(128) = 821$.

  • $\begingroup$ I hope you can find a similar logic explaining the first four result digits that you have chosen to ignore in this answer; you have fitted a second degree polynomial + an arbitrary degree of freedom (the reversal) to four two-digit data points. This is always possible, given whichever numbers on the left side, so if this were the intended solution, the puzzle wouldn't be very well constructed at all. $\endgroup$
    – Bass
    Aug 23, 2019 at 23:58
  • $\begingroup$ Yeah, this may be a mess compared to the first four digits, which are plainly just ROT13(gur zhygvcyvpngvba bs n naq o gura n naq p). So let's see what could be the intended answer. $\endgroup$
    – athin
    Aug 24, 2019 at 1:25

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