A credit card number has 5 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, and the fourth digit is 3 times the 5th digit. How many different credit card combinations are possible?

• 9
  third digit can be-> 7,8,9
  4th & 5th digit-> (3,1),(6,2),(9,3)
  total no of combination-> 3*3=9
• 11 years agoHelpfull: Yes(76) No(0)
• there will be 9 different combinations
  no's are
  12731
  12762
  12793
  12831
  12862
  12893
  12931
  12962
  12993
  
  
• 11 years agoHelpfull: Yes(41) No(0)
• This problem actually makes sense. We have 5 slots which we need to fill with different numbers. And we are given that the first and the second are 1 & 2 respectively.
  _1_ _2_ __ __ __
  
  The third digit is bigger than 6, therefore it could be either 7 or 8 or 9, since the range is [1.....9].
  the fourth digit is 3 times the 5th digit,so
  The fourth digit is divisible by 3, therefore it could be either 3 or 6 or 9.
  The fifth digit is 3 times the sixth, therefore it's multiple of 3, so in a given range it could be either 3 or 6 or 9. By the way, the sixth digit could be either 1 or 2 or 3, since the product of 5th and 6th can't be more than 9. So the reference to the 6th digit in the question is absolutely sensible.
  From this point, it looks like a simple combinatorics problem. We have 3 options in each of three decisions. So to calculate possible numbers we need to multiply 3 by 3 by 3. So the answer is 27
• 11 years agoHelpfull: Yes(10) No(28)
• 9
  12731 12831 12931
  12762 12862 12962
  12793 12893 12993
  
  
• 11 years agoHelpfull: Yes(4) No(1)
• 9
  Given that first two digits are 12 which are fixed
  3rd digit is greater than 6 so no. of possibilities are 7,8,9 i.e., 3
  4th digit = 3 * 5th digit
  => (9,3),(3,1),(6,2) i.e., 3
  => 3*3 = 9
  
  
• 11 years agoHelpfull: Yes(4) No(0)
• manohar gupta
  first two digit is fixed 12 then no of combination 1
  and third digid may be 7or8or9 then no of combination 3
  and 4th and 5th digid may be 31,62,93 because of fourth digid is 3 times of 5th digid so no of combination 3
  then total combination 1*3*1*3=9ans
• 11 years agoHelpfull: Yes(4) No(0)
• @parag : the credit card has only 5 digit.. but are using sixth digit ??
• 11 years agoHelpfull: Yes(3) No(4)
• 9 possible ways.
• 11 years agoHelpfull: Yes(2) No(3)
• 1st digit given=1
  2nd digit given=2
  3rd digit can be 7 or 8
  as 1 and 2 is already used so only 3*3
• 11 years agoHelpfull: Yes(1) No(7)
• repeation of digits is not allowed... how can u make a combination starting from 12 and ending with, a/c to me, 4th amd 5th digits are 9,3 only.... which results in only 2 possible combinations
• 11 years agoHelpfull: Yes(1) No(1)
• Answer is 3*3*3 = 27
• 11 years agoHelpfull: Yes(0) No(21)
• ANS: 72
  1 2 7 3 1 1 2 7 6 2 1 2 7 9 3
  1 2 8 6 2 1 2 8 3 1 1 2 8 9 3
  1 2 9 9 3 1 2 9 6 2 1 2 9 3 1
  
  now the above 9 patterns will be for 2,3; 3,4; 4,5; 5,6; 6,7; 7,8; 8,9 for 1st nd 2nd positions...
  so 9*8=72
• 11 years agoHelpfull: Yes(0) No(20)
• 12711,12,13,21,22,23,31,32,33
  12811,12,13,21,22,23,31,32,33
  12911,12,13,21,22,23,31,32,33
  == 27 combinations !!
• 11 years agoHelpfull: Yes(0) No(21)
• correct ans is 9
• 10 years agoHelpfull: Yes(0) No(0)