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My great-uncle Alfred has, for several years, voiced the opinion that the children of today have a much easier life than when he was young. Kids are mollycoddled by their parents, there is no more corporal punishment at school and even mathematics, he claims, has gotten easier. I am pursuing mathematics as a career which means that, at every opportunity, he will try to undermine my ability with a tricky riddle or maths problem to prove his point.

On our most recent interaction, great-uncle Alfred cornered me with a devious grin on his face.

"So, sonny," he said, "you think you're pretty smart, don't you, wanting to be a mathematician, eh?"

"Yes." I said, bluntly, not wanting to get into further discussion about it.

"Well," he smirked, "see if you can solve this formula."

He thrust a piece of paper into my hand with the following written on it

$\left[ \int_{2}^{sa} \frac{dx}{\ln x} \times AXCIX \times \left(\{ t | t \notin (-\infty,\infty)\}\times 2\right)\right] \times \left[ \frac{31}{12} \times \frac{lb}{in^2}0U \times 0.78571\ldots:1\right] \times \left[ \left(\min\{E[X], E[Y], E[Z]\} \gg 1 \right) \times \sqrt{4356} \times F\left(\frac{8 \clubsuit}{E} \right)\right] = \,\,\,?$

"The answer is an integer," he explained, "oh, and don't forget to get the correct units!"

Units, I thought to myself, I can't even make head nor tail of this mumbo-jumbo.

Has great-uncle Alfred lost his marbles or is there some sense to be made of this formula?

Can you detemine the solution with the correct units?
Can you explain the origin of each term in the formula?

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    $\begingroup$ It seems the answer requires some knowledge of Physics....and hope knowledge tag is enough...If not specifically physics tag is available. $\endgroup$
    – Mea Culpa Nay
    Commented Feb 28, 2018 at 16:19

2 Answers 2

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The first [...] indicates the following things (white text, select to see):

  • $\int_2^{sa}\frac{dx}{\ln x}$: $\color{white}{\textrm{Lisa}}$
  • $AXCIX$: $\color{white}{\textrm{Article 99}}$
  • $\left\{t\,\vert\notin(-\infty,+\infty)\right\}\times2$: $\color{white}{\textrm{Never Again}}$

all of which are

movies starring Jeffrey Tambor

The second [...] indicates the following things:

  • $\frac{31}{12}$: $\color{white}{\textrm{New Year's Eve}}$
  • $\frac{lb}{in^2}0U$: $\color{white}{\textrm{P.S.\ I Love You}}$
  • $0.78571...:1$: $\color{white}{\textrm{11:14}}$

all of which are

movies starring Hilary Swank.

The third [...] indicates the following things:

  • $\min\left\{E[X],E[Y],E[Z]\right\}\gg1$: $\color{white}{\textrm{Great Expectations}}$
  • $\sqrt{4356}$: $\color{white}{\textrm{Sixty Six}}$
  • $F\left(\frac{8\clubsuit}{E}\right)$: $\color{white}{\textrm{Fight Club}}$

all of which are

movies starring Helena Bonham Carter.

The solution is therefore

a movie starring all three: 55 Steps

which does indeed consist of a positive integer and a unit.

Credit where it's due:

Bass and I found a lot of the same things independently, but he got Fight Club which I didn't and, super-importantly, he identified the movie theme which I hadn't noticed at all. If you like this answer, go and upvote his too.

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  • $\begingroup$ Perfect. Well done to both of you. $\endgroup$
    – hexomino
    Commented Mar 1, 2018 at 19:14
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There sure is a lot going on in the formula, so here's a partial answer to get us started.

First of all, this looks like it might be some sort of mathematical rebus, or possibly even several of them.

At the beginning, the term $\int_{2}^{sa} \frac{dx}{\ln x}$ seems to be the definite logarithmic integral from 2 to "sa":

li(sa) - li(2), which is the same as the "offset logarithmic integral" Li(sa). (With a capital L.)

Can't make much anything of $AXCIX$, except maybe

A99 (roman numerals), which will probably need more interpretation once we know which direction we should go. Further interpreting the 99 as a decimal value for an ASCII character, we might guess "Ac", but that is definitely a bit of a stretch.

$\left(\{ t | t \notin (-\infty,\infty)\}\times 2\right)$ is a bit ambiguous, possibly algebraic notation for "t isn't on the open interval from negative infinity to positive infinity"; the "times 2" might be just to pluralise the word. The ambiguity rises from the parens, which can also mean ordered pairs in set theory, and possibly a couple of other things. A possible interpretation might be

NaN-$t$-s, chosen mostly because that letter combination fits English words better than many of the alternatives. Other options include "t-nan-ts" (tenants), "infinite t's" (infinities) and unreal-t's (unrealities).

$\frac{31}{12}$ looks a bit like a

date, specifically that of the new year's eve. So NYE or EVE, maybe?

$\frac{lb}{in^2}$ is

PSI, a pressure unit still used in some countries

0U needs work, but $0.78571\ldots:1$ seems to be an approximation of

11:14

$\min\{E[X], E[Y], E[Z]\} \gg 1 $ might imply that even if were the smallest one, the final term would still be a

Big Easy. ($E[Z]$ "is a lot bigger than" $1$)

Then,

$\sqrt{4356}$ is most likely

Route 66 (the square "route" of 4356 is 66)

and $F\frac{8 \clubsuit}{E}$ is most definitely

Fight Club

So there seems to be quite a few

Movie (or other pop culture) references

going on in here.

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  • $\begingroup$ Pretty sure the intention is that the logarithmic integral is taken to have 2 as its lower end, so that the first bit is just Li(sa) or "Lisa". I'd been wondering whether 0U was actually a mistake for 0L, which in C is a long integer constant, so you get NYE PSI LONG which has an "epsilon" in it. Probably not, though. I wondered whether the t bit was indicating the word "unreal", because one of the few things I could find for A99 was a weapon in a video game and there is a series of video games called Unreal. $\endgroup$
    – Gareth McCaughan ♦
    Commented Feb 28, 2018 at 20:30
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    $\begingroup$ I'd wondered about "route 66" and rejected it because that would have had to be $\sqrt{\sqrt{4356}}$. But it might well be right anyway. Excellent catch on the 11/14 and $8\clubsuit$ bits. $\endgroup$
    – Gareth McCaughan ♦
    Commented Feb 28, 2018 at 20:33
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    $\begingroup$ (Things I discovered while writing the previous comment that in retrospect should have been obvious: Googling for "latex clubs" is not an effective way to find out what runes are required to make that symbol.) $\endgroup$
    – Gareth McCaughan ♦
    Commented Feb 28, 2018 at 23:57
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    $\begingroup$ @GarethMcCaughan You presumably then switched to the more general search term "latex suits"? $\endgroup$
    – Bass
    Commented Mar 1, 2018 at 1:47
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    $\begingroup$ +1 You are definitely on the right track here and several guesses are indeed correct. There is one aspect which may push you in the wrong direction due to an oversight on my part. The answer to $\sqrt{4356}$ is not right. I love your guess (and wish that was it) but it is actually much simpler than that. Not all of the other guesses are correct but I feel this one will stick if I don't address it. $\endgroup$
    – hexomino
    Commented Mar 1, 2018 at 9:59

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