7
$\begingroup$

What word am I looking for? Try decipher the following:

((CZ18172612)⁺¹)=_____?

Edit; Gary S was very close, but here's a hint...

If numbers equals letters, then...

$\endgroup$
1
  • $\begingroup$ Is the word in any particular language? $\endgroup$ Commented Dec 17, 2019 at 13:15

2 Answers 2

2
$\begingroup$

It Is:

14 Mars or Mars 14

Because:

CZ = 4(26) + 1 = 41
And as per the explanation above, 18172612 = SRAM
Therefore, we got: 41 SRAM
After reversing it, we get Mars 14, or 14 Mars

As a conclusion:

I believe the word is "Pi", as March 14th is the National Pi Day

$\endgroup$
5
  • $\begingroup$ I've added spoiler markdown to your answer. $\endgroup$ Commented Dec 17, 2019 at 20:47
  • $\begingroup$ Yes you're correct, its Mars 14. And whats the word im looking for, you think? $\endgroup$ Commented Dec 17, 2019 at 21:13
  • $\begingroup$ Maybe rot13(Creuncf lbh ner gnyxvat nobhg gur Ntvyrag Uhzna 14 Zhygvcyr Nssvavgl Erzbiny Flfgrz NXN 14 Znef, fb gur jbeq lbh ner ybbxvat sbe vf Ntvyrag be cebgrva?) $\endgroup$
    – Nexevis
    Commented Dec 17, 2019 at 21:18
  • $\begingroup$ I've added my guess for the word too! $\endgroup$ Commented Dec 17, 2019 at 21:28
  • $\begingroup$ problem solved :) $\endgroup$ Commented Dec 17, 2019 at 21:33
7
$\begingroup$

Here's a guess:

DASRAM

Rationale:

It's not a common English word, but Google says it's a word, so I thought I would try it. I got this answer by applying the ⁺¹ to 1) each letter (i.e. C + 1 = D) and 2) each two-digit number as if it were a letter (i.e. 18 + 1 = 19 = S, 26 + 1 = 27 -> wraparound to 1 -> A).

D = C + 1
A = Z + 1 (wraparound)
S = (18 = R) + 1
R = (17 = Q) + 1
A = (26 = Z) + 1 (wraparound)
M = (12 = L) + 1

$\endgroup$
2
  • 1
    $\begingroup$ You're very very close... You just have to re-think what the meaning of the two letters in the beginning are.. $\endgroup$ Commented Dec 17, 2019 at 14:04
  • $\begingroup$ if rot13(Gur yrggref trg pbairegrq onpx gb ahzoref) then rot13( 427 FENZ ybbxf yvxr vg zvtug or n zbqry bs zbhagnva ovxr ) ... maybe? $\endgroup$
    – rm-vanda
    Commented Dec 17, 2019 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.