A through I are all binary variables (so they're either true of false).

How much is the binary number 0bABCDEFGH if all of these are true?

  • (A or B or C) and (D or E or F) and (G or H or I)
  • (C and D and F) or (A and I and G) or (B and E and H)
  • (B <=> C) and (A <=> F) and (I <=> B)
  • (I xor F) and (C xor D) and (E xor G)
  • A => F => I => B

(please also provide feedback how hard it was to solve)

  • 3
    $\begingroup$ You may want to clarify your question, as it stands, it doesn't make much sense and it doesn't look like much of a puzzle but rather a math problem. $\endgroup$
    – dcfyj
    Jan 19, 2017 at 20:40
  • $\begingroup$ I added "if all of these are true"? try to think about it for a bit and you will see what you need to solve. $\endgroup$
    – Hurda
    Jan 19, 2017 at 20:44
  • 2
    $\begingroup$ Did you deliberately say 0bABCDEFGH rather than 0bABCDEFGHI? $\endgroup$ Jan 19, 2017 at 21:57
  • $\begingroup$ 0bABCDEFGH is nice 8 bit number, originally this didn't have binary number mentioned in the question, nor explicitley stated that letters are all binary variables si this was kind of hidden hint, or confirmation $\endgroup$
    – Hurda
    Jan 20, 2017 at 18:59

1 Answer 1


It seems


is the solution, when

B, C, E, H and I are true, and A, D, F and G are false.


Rule 3: I, B, C have the same value (call it X), and A and F as well (call it Y).
Rule 4: I xor F must be true; this implies X != Y.
Rule 4: C xor D must be true; this implies D = Y.
In rule 2, (C and D and F) can never be true as C = X and D = F = Y.
In rule 2, (A and I and G) can never be true as A = Y and I = X.
Rule 2: So (B and E and H) must be true. This implies B = X = true.
Rule 4: E xor G implies G = false.

Note that

we haven't used the first and last conditions; therefore, they can be removed while keeping the solution.


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