Appendix I: Answers to puzzles from Part 3
1.
(8 / 2) * (2 +2) = 4 *4 = 16
Answer 2:
8 / (2 * (2 + 2)) = 8 / (2 * 4) = 8 /
8 = 1
2.
230 – (220 / 2) = 120
Answer 2:
(230 – 220) / 2 = 5
3.
(9 - 3) / (1/3 + 1) = 6 / 1 1/3 = 4.5
(actually it’s 4 2328/4656
more or less, but that’s just not worth the effort to figure out)
Answer 2:
((9 - 3) / 1) / (3 + 1) = (3 / 1) / 4
= 3/4
Answer 3:
(9 - (3 / 1)) / (3 + 1) = (9 – 3) / 4
= 6 / 4 = 1 1/2
Answer 4:
9 - (3 / 1/3) + 1 = (9 - 9) + 1 = 0 +
1 = 1
4.
(3 + 3) * (3 - 3 + 3) = 9 * 3 = 27
Answer 2:
3 + (3 * 3) - (3 + 3) = 3 + 9 - 6 = 3
Answer 3:
3 + (3 * (3 - 3 + 3)) = 3 + (3 * 3) =
3 + 9 = 12
Answer 4:
((3 + (3 * 3) – (3 + 3) = 3 + 9 - 6 =
6
Answer 5:
(((3 + 3) * 3) - 3) + 3 = (6 * 3) - 3
+ 3 = 18 - 3 + 3 = 18
5.
((1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +
1 + 1 + 1 + 1 + 1) * 0) + 1 = (14 * 0) + 1 = 0 + 1 = 1
Answer 2:
(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +
1 + 1 + 1 + 1 + 1) * (0 + 1) = 14 * 1 = 14
6.
20 + 20 + 20 = 60 (60 / 3 =20)
20 + 3 + 3 = 26 (26 - 20 = 6, 6 / 2 =
3)
3 + 6 + 6 = 15 (15 - 3 = 12, 12 /2
-6)
20 + 4 * 5
= ??
Answer 1:
(20 + 4) * 5 = 24 * 5 = 120
Answer 2:
20 + (4 * 5) = 20 + 20 = 40
If you insist on saying all the nests and bunches are
the same—and getting the wrong answers—here ya go:
Answer 1:
(20 + 3) * 6 = 23 * 6 = 138
Answer 2:
20 + (3 * 6) = 20 + 18 = 38
7.
10 + 10 + 10 = 30 (30 / 3 = 10)
5 + 10 + 10 = 25 (25 - 20 = 5)
7 + 5 + 5 = 17 (17 - 10 = 7)
7 + 5 * 10 = ?
Answer 1:
(7 + 5) * 10 = 12 * 10 = 120
Answer 2:
7 + (5 * 10) = 7 + 50 = 50
8.
8 + 8 + 8 = 24 (24 / 3 = 8)
6 + 6 + 8 = 20 (20 - 8 = 12, 12 / 2 =
6)
(6 * 2) - 2 = 10
(if Opera Mini is actually equal to 2)
8 + 6 / 2 = ??
Answer 1:
(8 + 6) / 2 = 14 / 2= 7
Answer 2:
8 + (6 / 2) = 8 + 3 = 11
If we group the third line as 6 * (2 - 2), it does not
equal 10; it equals 0. If try to solve for x, 6 * (x - x), good luck. I tried,
but couldn’t do it/didn’t have the patience to go beyond 6 = (10 – x) / x…If ya
wanna have a go at it, be my guest. Let me know what you come up with…
9.
10 + 10 + 10 = 30 (30 / 3 = 10)
10 + 5 + 5 = 20 (20 - 10 = 10, 10 / 2 = 5)
5 + 2 + 2 = 9 (9 - 5 = 4, 4 / 2 = 2)
5 + 2 * 10 = ?
Answer 1:
(5 + 2) * 10 = 7 + 10 = 70
Answer 2:
5 + (2 * 10) = 5+ 20 = 25
10.
Before you work your way through it
(yeah, like that’s going to happen; I didn’t…) I would like to lift two quotes
from it:
“Most of them present systems that
are quasi triangular, and hence, very very easy to solve.”
“Ok, the s integral is easy…”
So, no excuses for not getting this.
(But if you really don’t want to read this whole thing the short answer is 5Ï€/2,
and the hot dog = x.)
[NOTE: In pasting this into Word,
some of the symbols were lost, rendering the explanation below useless. I am only
putting it out there to show the complexity of the equation involved.]
And now, the blog entry in its
entirety.
nobody's blog
WhatsApp memes with hidden beauty!
November 8, 2018
Intro
I love internet memes. They provide a funny way to convey a
message or an idea. As a professor, I cherish all didactic mechanisms and
anything that easy the acquisition or fixation of ideias and concepts. In fact,
I find memes to be a very useful didactic tool. I might make a post only about
didactic memes in the future. Right now I want to talk about one specific meme
that, although not didactic at first glance, hides a cool and rather formal
math concept without most people realizing it (except for the author, I bet!).
The
fun…
So, here we go. This is the meme. Maybe you already received
it via WhatsApp (I did %-).
The beginning of the meme is, as you most probably know,
those famous “linear system of equations with emojis as variable names”. Fun
fact about those memes: Most of them present systems that are quasi triangular,
and hence, very very easy to solve. Anyways… In general, they give you 3 or 4
equations with right hand side values and asks you to compute a 4th or 5th
equation that is, in general, another linear combination of the variables.
In the case of the meme in this
post, the 4th equation is nothing like a linear combination of the values. In
fact, it is some weird looking complex mathematical aberration! This is where
the meme ends for most people. The funny in that meme is that nobody can solve
it. So, you suppose to send that to your annoying friend that keeps sending
regular linear combination meme to you, solving all of them to show that he is
a math genius.
Well… It turns out that the 4th equation is solvable!
Moreover, it involves the integral of a very famous function (sinc(x)) which has a very elegant solution for when the limits involves
infinity and zero. And yet more, that is despite the fact that it have no
closed form solution for arbitrary limits. Translating the emojis to variables, we arrive at the following integral
∫2b–a∞asin(x)cxdx
which, with numbers (solving the trivial linear system
of emojis), becomes
5∫0∞sin(x)xdx
The goal of this post them, is to solve this integral
using almost all steps as purely algebraic manipulations. We are also going to
use rules of calculus too, obviously. Moreover, we will have to go though some
complex numbers too (yes, we go that deep in a innocent internet meme). Lets
now solve the sinc function integral.
We start by acknowledging that sin(x)/x is a function
with an anti-derivative that is very hard to find. If only we had something to
relate easier functions to integrate (like exponentials)… But wait, we have!
The first exponential that we can exchange, is the sine function (as you
probably already guessed). So lets use Euler’s identity:
sin(s)=esj–e–sj2j
Although this is not a pure algebraic step, I will
save the Taylor expansion prove of this one, for it is a very, very known
relation.
The second one is the “jump of the cat” as we
say in my country. I found that in an awesome StackExchange answer[1]. It has
to do with the Laplace transform of the step function $u(t)$. In short, it
states that the Laplace of the step function is the inverse function. The
relation has a simple one line of proof as
L{u(t)}=∫0∞e–stdt=e–st–e–st–s∣∣∣∞0=1s
Hence
∫0∞e–stdt=1s
Holly butter ball! Now we can relate the inverse
function with the exponential. I wish I could use emojis on technical papers.
They convey so much of the authors wonder when some beautiful math trick is
used… %-)
Now lets change the names of the variables to
∫0∞sin(s)sds
which makes our original sinc integral to become
∫0∞∫0∞e–stesj–e–sj2jdtds
Ok, the s integral is
easy and here are the steps to solve it (it’s an exponential integral)
12j∫0∞∫0∞e–s(t–j)–e–s(t+j)dsdt12j∫0∞–e–s(t–j)t–j+e–s(t+j)t+j∣∣∣∞0dt12j∫0∞–e–∞(t–j)t–j–e–∞(t+j)t+j+e–0(t–j)t–j–e–0(t+j)t+jdt12j∫0∞1t–j–1t+jdt
Now we have a integral in t.
This one is a lot of fun. Lets start by separating the two integrals as
12j∫0∞1t–jdt–12j∫0∞1t+jdt
and do the following substitution.
u1=t–jdu1=dtu2=t+jdu2=dt
Now we foresee that we are going to have some trouble
with this infinity and the logs that will appear… Hum… Lets be careful (to not
say formal) and wrap that around a limit. We deal with the limit
later. Ok, so we have the following inverse integral (with is also simple to
solve inside the limit)
12jlima→∞∫–ja–j1u1du1–∫ja+j1u2du212jlima→∞ln(u1)|a–j–j–ln(u1)|a+jj12jlima→∞ln(a–j)–ln(–j)–ln(a+j)+ln(j)12jlima→∞ln(a–j)–ln(a+j)+ln(j)–ln(–j)12jlima→∞ln(a–j)–ln(a+j)+ln(j–j)ln(–1)2j+12jlima→∞ln(a–j)–ln(a+j)
The lets now investigate this peculiar limit (and
begin the fun part). We start by acknowledging that it is a complex-number
limit, hence not straight forward to evaluate. Then, we do the second “jump of
the cat” (small when compared with the Laplace one before) and change the limit
of infinity with a limit to zero as
lima→∞ln(a–j)–ln(a+j)lima→0ln(1a–j)–ln(1a+j)
Now we just do some algebraic steps as follows
lima→0ln(1–aja)–ln(1+aja)lima→0ln(1–aj)–ln(a)–ln(1+aj)+ln(a)lima→0ln(1–aj)–ln(1+aj)
And voilá! Put a=0 and we have out first part of the
integral
ln(1)–ln(1)=0
Now we have to figure what the heck is this negative
logarithm thing…
ln(–1)2j
To do that, lets make the cat jump again and write -1
as ejπ
now we have
ln(–1)2j=ln(ejÏ€)2j
which is just π2. %-)
Finally coming back to the original integral
we have
5∫0∞sin(x)xdx=5Ï€2
Which is the actual answer for the meme %-).
Conclusion
So, if you see this meme on your WhatsApp, feel free
to answer in a very “disdain” way: “Humpf! This is trivially equal to 5Ï€2!”. If
they doubt you, share this post with them %-).
I hope you like this post! Feel free to share
it or use any of this derivations on your own proofs! See you in the next post!
References:
[1] – Stack Exchange Mathematics – Integration of Sinc
function
11.
This may be may favorite of all the puzzles.
12 + 12 + 12 = 36 (36 / 3 = 12)
9 + 9 + 12 = 30 (30 - 12 = 18, 18 / 2
= 9)
3 + 3 + 9 = 15 (15 - 9 = 6, 6 / 2 =
3)
Smooth sailing so far, but then, WTF?
The fourth line has a couple of things going on. The
obvious one is that we are suddenly seeing symbols that haven’t appeared yet (yes,
they have), and second is that we have a doubled symbol. Looks like we are going
to have to dissect the symbols we’ve come across so far and hope for the best.
So, starting from the top, 12 is a combination of 1
hexagon and 6 triangles. Next 9 is a combination of 1 hexagon, 1 triangle, and
1 circle. 3 is made up of 1 triangle and 1 circle.
For the fun of it, let’s subtract the number of
triangles (6) in the 12 symbol from its value (12). That leaves us with 6. If
we assign a value of 6 to the hexagon and 1 to each of the triangles, they add
up to our happy number 12. Will this hold up? Let us see.
If the hexagon in the 9 symbol has a value of 6, and
the triangle has a value of 1, if we subtract them from 9, it leaves us with 2.
So, it appears the value of a circle is 2.
Next comes the 3 symbol. Adding a triangle (1) and a
circle (2), does indeed give us 3. It all seems to work out. We are ready for
the next line.
(2 * 2) + (2 * 2) + 6 = 4 + 4 + 6 =
14 (14 - 6 = 8, 8 / 2 = 4)
Deciphering the next line, we have 1 circle (2) with 5
triangles (5 * 1 = 5) making 7, and 1 hexagon (6) with 1 triangle (1)—what do
you know—also making 7. You can write seven (7) two different ways? So, we
finally have the next line.
7 + 7 * 2 = ?
Answer 1:
(7 + 7) * 2 = 14 * 2 = 28
Answer 2:
7 + (7 * 2) = 7 + 14 = 21
For extra credit, try “writing” the answers using the
symbols given. Are there enough? Will have to invent some? I don’t know. I
didn’t try. But it certainly seems like you could.
12.
Nope. No answer here. If you were paying attention,
you would know that I already went through this one.
13.
You’re
gonna kick yourself if you didn’t get this one. None of the equations is
“wrong” per se, although multiple answers are possible. But you don’t even need
them to answer the question. Cover them up and look at the question again. Now
do you see a mistake?
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