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 $$3+\sum_{k=0}^\infty \dfrac{1}{k!} = 8$$.
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$\begingroup$ I don't see any variables. What are we solving for? $\endgroup$– LeppyR64Jun 1, 2016 at 17:08
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1$\begingroup$ You can't really prove it unless you prove that $3 + e = 8$. Because the sum evaluates to $e$. $\endgroup$– user88Jun 1, 2016 at 17:10
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6$\begingroup$ e is the fifth letter of the alphabet $\endgroup$– Jared GoguenJun 1, 2016 at 17:36
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1$\begingroup$ which alphabet? $\endgroup$– Jeremy ArgentJun 1, 2016 at 17:54
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1$\begingroup$ @JaredGoguen You should post that as an answer $\endgroup$– aebabisJun 1, 2016 at 22:36
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3 Answers
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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:
$3 + E = E + 3 \to E\hspace{0.2em}3 \to E\hspace{-0.1em}3 \to E\hspace{-0.3em}3 \to 8$
answered Jun 1, 2016 at 17:15
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$\begingroup$ 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. $\endgroup$ Jun 1, 2016 at 20:40
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$\begingroup$ @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. $\endgroup$– user88Jun 1, 2016 at 20:53
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$\begingroup$ @JoeZ. Don't know why it played out that way, but I did see it as it was happening. $\endgroup$ Jun 1, 2016 at 21:04
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$\begingroup$ @JoeZ. I do feel badly about that. I'll put your name is so all glory to Joe Z. $\endgroup$ Jun 1, 2016 at 21:18
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$\begingroup$ The last transformation is a bit of a stretch imo. $\endgroup$– OlegJun 1, 2016 at 21:21
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Did he write it on the blackboard perchance? If so
define $l=\frac5e$
now
$$3+\sum_{k=0}^{\infty}\frac{l}{k!}=8$$
Another possibility is
To read the concatenation in Russian:
$$3+\sum_{k=0}^{\infty}\frac{1}{k!}=3+e=\text{зе}=\text{ж}=8$$
Since зе is pronounced the same as ж, the 8th letter in the alphabet.
Although this would probably be better asked as:
$$3\sum_{k=0}^{\infty}\frac{1}{k!}=8$$
answered Jun 2, 2016 at 11:45
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$\begingroup$ Umm... No, it is not pronounced the same, it's more like "ze" vs. "zh" $\endgroup$– user10531Jun 21, 2016 at 19:40
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$\begingroup$ @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. $\endgroup$ Jun 21, 2016 at 20:00
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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.
