Denote by M & AMP; N Diffex conjunction of non-negative integers m and n.

so, for example, 14 & amp; 5 = 1110 & amp; 0101 = 0100 = 4. For which smallest non-negative integer A formula

x & amp; 29 ≠ 0 → (X & amp; 17 = 0 → X & amp; A ≠ 0)

identically true (i.e. takes value 1 at any non-negative whole value of the variable x)?

wrote the function of this conjunction and implication, then the function of the extinguishing of all x to 100 with any A, which will return False, if at least one value has not passed.

def conc (n, k):
  n = bin (n) [2:]
  k = bin (k) [2:]
  s = ''
  IF Len (n)! = Len (k):
    IF LEN (N) & GT; Len (k):
      k = '0' * (Len (n) - len (k)) + k
    ELSE:
      n = '0' * (len (k) - len (n)) + n
  For i in Range (Len (N)):
    IF N [i] == K [i]:
      S + = '1'
    ELSE:
      S + = '0'
  RETURN INT (S, 2)
Def impl (X, Y): Return Not X OR Y
DEF F (A):
  For X in Range (10,000):
    IF Not impl (CONC (X, 29)! = 0, impl (CONC (X, 17) == 0, CONC (X, A)! = 0)):
      Return False.
  Return A.
For i in Range (100):
  IF F (i)! = False: Print (I)

Conjunction and implication functions like the correct, but when executing the program, it gives all values ​​from 2 to 100. In response to the task number 12. Where I was wrong, I will fight over the challenge, but I can’t understand ??

Answer 1, Authority 100%

if n [i] == k [i]:
    S + = '1'
  ELSE:
    S + = '0'

This is not Bitched and (conjunction), it is necessary that both discharge are not only equal to each other, but also equal to 1 at the same time, so it will be right:

ife n [i] == k [i] == '1':
    S + = '1'
  ELSE:
    S + = '0'

After this correction code output:

12
13
fourteen
15
28.
29.
thirty
31.
44.
45.
46.
47.
60.
61.
62.
63.
76.
77.
78.
79.
92.
93.
94.
95.