2
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My friend gave me this sequence of numbers:


3=11


6=12


11=13


14=14


361114=A


A=?


Which number can replace the question mark?

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4
  • $\begingroup$ This is the original! $\endgroup$
    – Tanton
    Aug 16, 2019 at 9:28
  • $\begingroup$ Does A change the answer? $\endgroup$
    – Abbas
    Aug 16, 2019 at 9:37
  • $\begingroup$ A It doesn't make any sense $\endgroup$
    – Tanton
    Aug 16, 2019 at 10:34
  • 4
    $\begingroup$ Could the last two lines have been replaced with 361114=? instead of adding a variable into the mix? $\endgroup$ Aug 16, 2019 at 13:17

4 Answers 4

2
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New Solution:

10. A (hexadecimal) = 10

Another possible Solution could be:

1. Since A is the first letter in the alphabet

If it's

11121314

then it's too easy.

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13
  • $\begingroup$ A is just a medium for transformation! $\endgroup$
    – Tanton
    Aug 16, 2019 at 9:49
  • $\begingroup$ I'm thinking the last one here is the most applicable, and probably the answer to the question. Can you confirm/deny @Tanton? $\endgroup$
    – Birjolaxew
    Aug 16, 2019 at 10:41
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    $\begingroup$ @Tanton maybe you should discuss this with your friend before you accept this answer $\endgroup$
    – Adam
    Aug 16, 2019 at 11:14
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    $\begingroup$ This answer currently says that "A (hexadecimal) = 11" but in hexadecimal A means 10. No part of it has any actual explanation that I can make any sense of it. @Tanton I really don't think this should be accepted. $\endgroup$
    – Gareth McCaughan
    Aug 16, 2019 at 21:23
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    $\begingroup$ I literally cannot understand what you're saying with this answer. Both "new" and "another possible" solutions ignore the entirety of the puzzle save its final entry and focus on the letter, whose specific inclusion may indeed be relevant somehow but certainly shouldn't be the be-all-end-all of the puzzle. This is supposed to be a sequence of numbers; you've discarded the sequence entirely and just proposed a substitution of number for variable. How is this a solution to the sequence? @Tanton how is this possibly a satisfying answer? $\endgroup$
    – Rubio
    Aug 17, 2019 at 15:37
7
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I also have a solution

123468

because,

Think of the number to the left of the sequence as octal

Like

3(octal)= 11(binary)
6(octal)= 12(quanternary)
11(octal)= 13(Senary)
14(octal)= 14(octal)

So

361114(octal)= 123468(decimal)

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3
  • $\begingroup$ My answer is based on the fact that A has no effect. Because A=123468 is substituted into the next formula, I can't continue to convert. $\endgroup$
    – sypicky
    Aug 16, 2019 at 14:41
  • 2
    $\begingroup$ What determines the bases used on the right-hand sides? $\endgroup$
    – Gareth McCaughan
    Aug 16, 2019 at 21:25
  • 2
    $\begingroup$ On the right hand side is a series of even numbers, and adds 2.I think it's a rule. $\endgroup$
    – sypicky
    Aug 17, 2019 at 0:05
4
$\begingroup$

An alternate solution is:

24024

because,

generally, when you have variable names next to each other (e.g. $xy$) that would mean that you are multiplying their values (e.g. $x\times y$).

Therefore,

you would multiply $11\times12\times13\times14=24024$

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9
  • 1
    $\begingroup$ Also a valid theory, It could be your answer is the one OP's friend was looking for! $\endgroup$
    – Abbas
    Aug 16, 2019 at 14:15
  • $\begingroup$ How does this account for things like "11=13" in the question? $\endgroup$
    – Gareth McCaughan
    Aug 16, 2019 at 21:24
  • $\begingroup$ Maybe I'm just slow today... why did you select those numbers to multiply? $\endgroup$ Aug 16, 2019 at 21:33
  • $\begingroup$ I don't think multipliers can be omitted between numbers. One example is 6÷2 (1 + 2). $\endgroup$
    – sypicky
    Aug 17, 2019 at 0:18
  • $\begingroup$ @sypicky - That's true, but neither can you say $3=11$... This is assuming the numbers on the left are variables. $\endgroup$ Aug 17, 2019 at 1:46
0
$\begingroup$

Based on the last but one line, A is

361114

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2
  • $\begingroup$ Well if A is this number then why is 3=11 or 6=12 or 11=13? $\endgroup$
    – Adam
    Aug 16, 2019 at 18:48
  • $\begingroup$ As, I have just seen Voldermort's 1st comment to the OP and up-voted it. $\endgroup$ Aug 17, 2019 at 6:34

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