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I have been trying to contact my son Mark for past 4 days, I hope he is alright. My 17 year old son went to attend "Mathematics Summer Camp", I am really eager to know about the camp, his friends, stories of his camp. Unfortunately, I am unable to contact him through his phone. I also received a letter from Mark, I don't really know what it means, I am really worried as if he's trying to say something through the letter. enter image description here

Can you figure out the letter from Mark?

Hint 1:

11/7 = Pi/2

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  • 1
    $\begingroup$ Is the last character on the man's face a ¿ or an i? $\endgroup$ Commented May 17, 2022 at 21:13
  • 1
    $\begingroup$ That last character is i. The number/alphabets in order are, 13, 19i, 2, 2, 11/7, 64, 32, 100, 19.7, in case of any confusion over the text $\endgroup$
    – I'm Nobody
    Commented May 17, 2022 at 21:29
  • $\begingroup$ Does "i" refer to the square root of minus one? $\endgroup$
    – DontMindMe
    Commented May 18, 2022 at 0:43
  • $\begingroup$ rot13: BZT! Vzntvanel havg. Ybtnevguzvp shapgvba. Sybbe shapgvba. Ebbg shapgvba. Guvf vf gbb shaal. $\endgroup$
    – Galen
    Commented May 18, 2022 at 21:00
  • $\begingroup$ @I'mNobody we need more hints... eish $\endgroup$ Commented May 19, 2022 at 5:59

2 Answers 2

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Very close to solving but can't figure out the last step. If someone wants to work off this feel free.

The Face Picture:

No eye means remove the $i$ in some way.
We could just cut the $i$ out of the expression and get $19+13=32$, or we could cut out the $i$ term and get $19$. Or we could multiply by the conjugate $(13+19i)(13-19i)=530$
This pic probably means 32, 19 or 530.

Edit 1: I read the problem wrong, thank you to @Galen to pointing this out.

Cutting out the $i$ term means it would be 13, not 19. Before I read it as $19+13i$ instead of $13+19i$ hence the mistake.

The Log Picture:

Log Base 2

logb2(2)+ logb2(64) + logb2(32) = 1 + 6 + 5 = 12.
OR logb2(2) * logb2(64) * logb2(32) = 1 * 6 * 5 = 30.
OR, incorporating the 11/7, which is above 64, it could be:
11/7 * (logb2(2) + logb2(64)) + logb2(32) = 11 + 5 = 16.

Edit 1: thank you @ribs2spare for pointing this out

11/7 is on a sign. So it means SINE(11/7). 22/7 is the common approximation for pi so $\sin(\frac{11}{7})=\sin(\frac{\pi}{2})=1$. So its basically 1 times the log stuff or 1 plus the log stuff so either 12 or 13.

The tree picture:

The 100 is under roots, so root 100 = 10

The room picture:

The floor function means round down. 19.7-->19

Combining them (or at least trying to):

My first guess was this is some kind of convert numbers to letters cypher. Using the numbers 19, 16, 10, 19 gives SPJS. I tried rot## but didn't get anything interesting.
Now trying to round other options for the face picture: KPJS and GPJS
My next strategy is concatenating the numbers and doing base 26. These are my results: BSQDXWL, BPYESH, CSJVMH.
Multiplying instead of Concatenating. Results: DNRLG, FNXO, DHLO

Edit 1: Alright now that I've cleared up some mistakes I'm gonna give this another go.
! The numbers are now 13, 12, 10, 19 --> MLJS or 13, 13, 10, 19 --> MMJS.

Still a bit lost.

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  • $\begingroup$ The problem is not your last step, the numbers you have derived are not correct. You are definitely on the right track. $\endgroup$
    – I'm Nobody
    Commented May 18, 2022 at 6:07
  • $\begingroup$ rot13(Gur snpg gung 11/7 vf ba n "fvta" yrnqf zr gb oryvrir gung fva bs 11/7 fubhyq or cneg bs guvf) $\endgroup$
    – ribs2spare
    Commented May 18, 2022 at 12:14
  • $\begingroup$ rot13: Gur "ab rlr" cneg zvtug zrna 'gnxr gur erny cneg'. $\endgroup$
    – Galen
    Commented May 18, 2022 at 21:02
  • $\begingroup$ rot13: Be znlor "ab rlr" pbhyq or whfg gnxvat gur fhz bs 13+19=32, ohg zl vaghvgvba vf gung gur erny-cneg shapgvba vf zber yvxryl. $\endgroup$
    – Galen
    Commented May 18, 2022 at 21:07
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    $\begingroup$ @ribs2spare Thank you for the help. I added ur contribution and put your name down to give credit. $\endgroup$
    – Ankit
    Commented May 19, 2022 at 2:16
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My answer, credit, @Ankit

The face:

No i typically means the real part which is 13

The log:

log2 of 2, 64, and 32 are 1, 6, and 5

The sign:

Sine of 11/7 is 1

The tree:

100 is under roots. We can infer that this is square root, yielding 10

The floor:

The floor of the decimal is 19

We do not:

Combine 1 6 5 from the log, because there are, quote, 7 numbers

Next step:

We use A1Z26 to get MAFEAJS from 13 1 6 5 1 10 19

Why this seems on the right track:

There are three vowels out of the total of seven letters, which can make a decent word

However:

There isn't any anagrams for it. It could've contain "SAFE", but it would've been "SAFE JAM"

Wait:

What about differences? For instance, 13-1=12. You get 12 5 1 4 9 9 which is for leadii

Why this seems to be good:

There seems to be still enough vowels and we have a double letter, useful for a caesar cipher

But:

No useful stuff ocurr.

Going back:

The sign is on the log, so maybe we apply log to 1? We get 0 mapping to Z (warping around)

So we get:

13 1 6 5 0 10 19 or MAFEZJS

This seems wrong because:

We have too less vowels

Perhaps meaning:

Caesar cipher

or

Virgenere cipher

But:

Both fail

I think:

We still have a problem with the numbers

Edit:

Nobody ( :-) ) pointed out that the SAFE JAM should be I AM SAFE

So:

The 10 is wrong and should be a 9

So:

The root was sort of a wrong interpretation

So:

We have to link 100 to 9

How?

It could have been digit root, but I don't see why. Maybe later.

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  • $\begingroup$ Rot13(Lbh'er irel pybfr gb gur nafjre, Qba'g lbh guvax "FNSR WNZ" jvyy orpbzr n zrnavatshy cuenfr nsgre lbh fjnc W jvgu V?) $\endgroup$
    – I'm Nobody
    Commented May 19, 2022 at 6:14
  • $\begingroup$ Need some help. $\endgroup$ Commented May 19, 2022 at 6:27
  • $\begingroup$ Znlor fbzrguvat jvgu v.v.19 = -19 orvat nqqrq gb gur 100 tvivat 81. V qba'g xabj ubj gb whfgvsl gung jvgu gur cvpgher gubhtu. Fgnegf gb srry yvxr uggcf://ra.jvxvcrqvn.bet/jvxv/Ntr_bs_gur_pncgnva $\endgroup$
    – Dr Xorile
    Commented May 19, 2022 at 23:16
  • $\begingroup$ Please give credit... I don't mind you working off my answer but please give credit because half of your answer was already in mine. $\endgroup$
    – Ankit
    Commented May 22, 2022 at 1:37
  • $\begingroup$ Oh OK @Ankit IHATE15CHAR $\endgroup$ Commented May 22, 2022 at 4:05

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