# Find pattern for new equation in number sequence

I saw this puzzle on this address.

$$3+5+7=152181$$

$$4+5+6=202461$$

$$3+4+7=122172$$

$$9+4+5=364518$$

$$8+6+8=?$$

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

Surprisingly, the last $$2$$ digits will be

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

Because

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 $$rev(8^2 \times (8-6)) = rev(128) = 821$$.

• 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. – Bass Aug 23 '19 at 23:58
• 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. – athin Aug 24 '19 at 1:25