The recent spat of 22+4 puzzles took me back to 1974, when my high school math teacher (and cross-country coach), Dr James Quinlan, asked us to solve $$3+\sum_{k=0}^\infty \dfrac{1}{k!} = 8$$.

  • $\begingroup$ I don't see any variables. What are we solving for? $\endgroup$ – LeppyR64 Jun 1 '16 at 17:08
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    $\begingroup$ You can't really prove it unless you prove that $3 + e = 8$. Because the sum evaluates to $e$. $\endgroup$ – user88 Jun 1 '16 at 17:10
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    $\begingroup$ e is the fifth letter of the alphabet $\endgroup$ – Jared Goguen Jun 1 '16 at 17:36
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    $\begingroup$ which alphabet? $\endgroup$ – Jeremy Argent Jun 1 '16 at 17:54
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    $\begingroup$ @JaredGoguen You should post that as an answer $\endgroup$ – aebabis Jun 1 '16 at 22:36

This was just posted by Joe Z., quickly downvoted, and then deleted. I don't understand why. It seemed like a legitimate and puzzle-y answer to me. I am posting a more complete version now so people can tell me if I'm wrong.

The equation evaluates to be

$3+e=8$     because     $\sum_{k=0}^\infty \dfrac{1}{k!} = e$     by definition.

If you capitalize that and squish them together, it looks like this:

MATH $3 + E = E + 3 \to E\hspace{0.2em}3 \to E\hspace{-0.1em}3 \to E\hspace{-0.3em}3 \to 8$

  • $\begingroup$ I would up-vote 5 times if I could. A somewhat more elegant, imo, solution exists. I'll give it a few days to see if it gets posted. $\endgroup$ – Jeremy Argent Jun 1 '16 at 20:40
  • $\begingroup$ @EngineerToast I posted it as a sort of joke answer that still kinda worked, and when I saw that people immediately didn't like it (2 downvotes in 1 minute is a pretty strong message), I deleted it. $\endgroup$ – user88 Jun 1 '16 at 20:53
  • $\begingroup$ @JoeZ. Don't know why it played out that way, but I did see it as it was happening. $\endgroup$ – Jeremy Argent Jun 1 '16 at 21:04
  • $\begingroup$ @JoeZ. I do feel badly about that. I'll put your name is so all glory to Joe Z. $\endgroup$ – Engineer Toast Jun 1 '16 at 21:18
  • $\begingroup$ The last transformation is a bit of a stretch imo. $\endgroup$ – Oleg Jun 1 '16 at 21:21

Did he write it on the blackboard perchance? If so

define $l=\frac5e$

Another possibility is

To read the concatenation in Russian:
Since зе is pronounced the same as ж, the 8th letter in the alphabet.

Although this would probably be better asked as:

  • $\begingroup$ Umm... No, it is not pronounced the same, it's more like "ze" vs. "zh" $\endgroup$ – user10531 Jun 21 '16 at 19:40
  • $\begingroup$ @Akiiino thanks for letting me know! At the time I listened to pronunciation on the web and they sounded the same, but I am not Russian and have never studied it. $\endgroup$ – Jonathan Allan Jun 21 '16 at 20:00

As noted in other answers, the equation simplifies to:

$3+\sum_{k=0}^\infty \dfrac{1}{k!}=3+e$

Using the trivial letter number substitution cipher leads to:

$3+5=8$ since $e$ is the fifth letter of the English alphabet.


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