Problem of the Day
A new programming or logic puzzle every Mon-Fri

For the numbers 1-100 print out "true" if the 2nd most least significant bit is set to 1 for that number. If not print out "false".

Comments:

• Anonymous - 10 years, 4 months ago

  if (n Mod 4 >= 2) true

  else false

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• Cha0z97 - 10 years, 4 months ago

  #include <stdio.h>

  int main() { int i; for (i=1;i<=100;i++) { i&2? printf("true\n") : printf("false\n"); } }

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• Johnathan - 10 years, 4 months ago

  for i in range(1, 100):
      if format(i, '#010b')[8:-1] == '1':
          print 'true'
      else:
          print 'false'
  raw_input('')
  

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• Johnathan - 10 years, 4 months ago

  I guess format(i, '08b')[6:-1] would have made more sense. I really need to read over my programs more before I submit them. Oh well.

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• Nick Krichevsky - 10 years, 4 months ago

  for i in range(1,101):
      if bin(i)[-1]==1:
          print 'true'
      else:
          print 'false'
  

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• (Bernardo Botelho)[http://github.com/bzxbot] - 10 years, 4 months ago

  Ruby:

  (1..100).each {|n| puts n[1] == 1}

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• bzxbot - 10 years, 4 months ago

  Ruby:

  (1..100).each {|n| puts n[1] == 1}

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• Anonymous - 10 years, 3 months ago

  Don’t overcomplicate things, just do a bitwise and

  for i in range(1, 101):
      print("true" if i & 2 else "false”)
  

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