A credit card number has 6 digits (between 1 to 9). The firs

A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?
A. 27
B. 36
C. 72
D. 112
E. 422

{1}{2}{greater than 6}{divisible by 3}{3x}{x}:

The third digit can take 3 values: 7, 8, or 9.
The fourth digit can take 3 values: 3, 6, or 9.
The fifth and sixth digits can take 3 values: 31, 62, or 93.

Total = 3*3*3 = 27.

Answer: A.

Hope it helps.
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Nice. That is something to look out for on the test. If there is a dependency of one variable on another you treat the two variables as one variable when figuring out combinations as well as odds.

Hey Guys - I tried to do this problem but not sure what is the OA for this one, any clue?

The way I did it was:

Nevermind about the first 2 digits, they are fixed as 1 and 2 respectively.
Then for the 3rd digit, has to be greater than 6, so could be (7,8,9) any of those 3 choices.
For the 4th has to be divisible by 3 so could only be (3,6,9), 3 choices more.
For the 5th and the 6th digit, first note that the 5th digit is 3 times the 6th digit so then the 6th digit could only be (1,2,3) since the 5th digit cannot exceed 9.
So there are 3 choices for the 6th digit and only 1 choice for the 5th digit (because it depends entirely on what the 6th digit is)

So IMO Answer is (A) 27

Please let me know if i'm right on this one.

I have a doubt on this question. It says between 1 to 9 so do we have to consider 1 and 9 because it thought we have to consider only 2,3,4,5,6,7,8 as the total number available. so the 3rd place can be filled with 2 ways (7 or 8) then the 4th place can be filled with 2 ways (3 or 6). I am not understanding how to fill the 5th and the 6th place because if you consider any value for the 6th place you get a 2 digit number for the 5th place right?? Am i wrong in my steps?? Kindly help

Between 1 and 9(include 1 and 9 in this case)
Now, considering that information,

The third digit is 7,8 or 9.(3 combinations)
Fourth digit which is divisible by 3, has to be 3,6, or 9(3 combinations)

Coming to the fifth and sixth digits,
Since fifth digit's value is thrice sixth's digit value
Again there are 3 combinations for the combination of fifth and sixth digits

5th Digit ---- 6th digit
----3---------------1-----
----6---------------2-----
----9---------------3-----

They are 3 combinations possible for the combination
All in all, the total ways the credit card could be formed are 3*3*3 = 27 ways(Option A)

Hope that helps!

Hi,
So what I don't understand in this problem is that since for the 3rd digit we are choosing (7,8,9) that's 3 options how is that we have 3 options for the 4th digit (divisible by 3 ) . If I had chosen 9 as my 3 rd digit how can I have that as an option for my 4th digit.

Hi Superg8

Read the questions properly
A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth one can be equally divided by 3 and the fifth digit is 3 times bigger than the sixth one. How many credit cards can be made?

It has not been given that the digits cannot repeat.
So having 9 as both the 3rd and 4th digits is possible. Both 129993 and 127362 are valid credit card numbers.

Hope that clears your confusion!

Why 4 th digit cannot be zero as it is divisble by 3 .Answer will change accordingly as there will be 4 option for filling the 4 th digit (0,3,6,9)

The question mentions that card has 6 digits (between 1 and 9)

Posted from my mobile device

My doubt is why zero is not considered for 5th and 6th place ?

Posted from my mobile device

The question gives you that the digits are "between 1 to 9".
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Hi, I've been solving quite a lot of such questions and I still haven't been able to come up with a definitive answer to this question: "If nothing is mentioned, should we just assume that repetition is allowed?"
I think, yes... right?

Hi. Its mentioned in the problem of taking the digits from 1-9. If it wasn't, would be consider a 0? Is 0 considered to be divisible by 3?

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