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Imagine you are on a spaceship. To activate this machine, you have to enter a code with these given clues.

1110C1742

0110F1230

1001B1126

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You have to enter the fourth code using one of the given codes. Which code suits the logic above?

  1. 10110E2733

  2. 11001Z51146

  3. 1111A3240

  4. 11011C3036

Hints:

  1. Simple addition and multiplication with 2

  2. 0100100101110010, what is it?

  3. What value does the slanted number holds?

  4. 64×1+8×2+3=83

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2
  • $\begingroup$ Hint 2 when translated leads to rot13(ve). Hmmmmm $\endgroup$
    – user71418
    Commented Oct 13, 2020 at 8:39
  • $\begingroup$ It is just a random binary code. I am just telling you that it is a binary code, it doesn't mean anything $\endgroup$
    – 00xxqhxx00
    Commented Oct 13, 2020 at 13:07

1 Answer 1

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The next code is

11001Z51146

The pattern is

bin + A1Z26 = dec; dec * 2 = oct

With the example of 1110C1742:
11102 = 1410
C = 3
14 + 3 = 17
428 = 3410 = 17 * 2

And with the answer, 11001Z51146:
110012 = 2510
Z = 26
25 + 26 = 51
1468 = 10210 = 51 * 2

If this is the correct rule, the answer is the only possible one because

The octal numbers of the of the other options do not equal x*2, but x

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