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A credit card number has 6 digits (between 1 to 9). The firs [#permalink]
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Updated on: 12 Sep 2013, 07:47
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64% (01:23) correct 36% (01:51) wrong based on 478 sessions
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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
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Originally posted by Value on 19 Jun 2008, 04:30.
Last edited by Bunuel on 12 Sep 2013, 07:47, edited 2 times in total.
Edited the question and added the OA.



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Re: PSCombinations [#permalink]
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19 Jun 2008, 06:13
Value wrote: 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?
(a) 27. (b) 36. (c) 72. (d) 112. (e) 422.
Please explain your answer. IMHO Answer should be 81, 3 possibilities for 3rd digit(7,8,9) 3 possibilities for 4th digit(3,6,9) 3 possibilities for 5th digit(3,6,9) 3 possibilities for 6th digit(1,2,3) 3^4=81 Am I doing something wrong?



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Re: PSCombinations [#permalink]
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19 Jun 2008, 09:23
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.



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Re: PSCombinations [#permalink]
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21 Jun 2008, 07:28
Value wrote: 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?
(a) 27. (b) 36. (c) 72. (d) 112. (e) 422.
Please explain your answer. What do you mean by "The first two digits are 12 in that order"? I got the 3rd to the 6th digit right, but i didn't understand the first two digits...



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Re: PSCombinations [#permalink]
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21 Jun 2008, 15:06
It means that the first two digits will always be 12. They are fixed.



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Re: PSCombinations [#permalink]
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22 Jun 2008, 08:45
Well, if the first two digits are fixed then entire solultion would be different. Now suppose if 1st and 2nd digits are fixed, they are 1 and 2 Then 1st digit=1 2nd digit=2 6th digit= 6 5th digit= 9 4th digit = 6 3rd digit= 4/5/7/8 Now we can interchange 1st digit and 6th digit so total cards= 4*2=8.
Now only variable is 3rd digit, rest all are fixed. Am I doing something wrong?



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Re: A credit card number has 6 digits (between 1 to 9). The [#permalink]
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12 Sep 2013, 05:52
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.



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Re: A credit card number has 6 digits (between 1 to 9). The [#permalink]
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12 Sep 2013, 07:50
jlgdr wrote: 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. 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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Re: A credit card number has 6 digits (between 1 to 9). The [#permalink]
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11 Nov 2013, 15:18
jlgdr wrote: 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. Good catch. Digits 5 & 6 are dependent on each other and are treated as 1 combination.



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Re: PSCombinations [#permalink]
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23 Nov 2013, 06:14
alpha_plus_gamma wrote: Value wrote: 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?
(a) 27. (b) 36. (c) 72. (d) 112. (e) 422.
Please explain your answer. IMHO Answer should be 81, 3 possibilities for 3rd digit(7,8,9) 3 possibilities for 4th digit(3,6,9) 3 possibilities for 5th digit(3,6,9) 3 possibilities for 6th digit(1,2,3) 3^4=81 Am I doing something wrong? 5th and 6th digit are dependant so only count as 3 different combinations



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Re: A credit card number has 6 digits (between 1 to 9). The firs [#permalink]
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13 Dec 2013, 00:08
(1*1*3*3*1*1) *3=27 ways of picking digits from 1 to 6, then multiplying by 3 as there are 3 ways to pick 2 last digits (it can be 1 and 3, 2 and 6, or 3 and 9)



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Re: A credit card number has 6 digits (between 1 to 9). The firs [#permalink]
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13 Dec 2013, 08:10
can someone please explain/solve this question, i have been trying to solve it from the past four hours and now posting here to see if someone can answer. Q How many 4letter combinations are there of the letter in each word? a ONOWAY bOSBORNE c OUTLOOK
Ansa11, b25 c15



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Re: A credit card number has 6 digits (between 1 to 9). The firs [#permalink]
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02 Aug 2015, 00:44
alpha_plus_gamma wrote: Value wrote: 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?
(a) 27. (b) 36. (c) 72. (d) 112. (e) 422.
Please explain your answer. IMHO Answer should be 81, 3 possibilities for 3rd digit(7,8,9) 3 possibilities for 4th digit(3,6,9) 3 possibilities for 5th digit(3,6,9) 3 possibilities for 6th digit(1,2,3) 3^4=81 Am I doing something wrong? alpha_plus_gamma as you said there are 3*3 possibilities combining 3rd and 4th digit but when it comes to 5th digit it has a condition with 6th digit and there are only 3 possibilities for 6th digit which, when one is selected gives 5th digit only one option to comply with the given condition .i.e. if 6th digit is 1  5th is 3,if 6th is 2  5th is 6 and if 6th digit is 3 – 5th is 9. So in total there are only 3 possibilities. 3*3*3=27. I know its very old post but i felt like replying.



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Re: A credit card number has 6 digits (between 1 to 9). The firs [#permalink]
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25 Nov 2015, 19:11
I don't agree with the OA. 0 is divisible by 3, thus we can have: 7,8,9 in the 3rd digit 0,3,6,9  in the 4th digit the 6th digit can be 1,2,3 the 5th digit can be 3,6,9
total, we have 1x1x3x4x3x3 options.
we can eliminate D and E because both numbers are not divisible by 3.



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Re: A credit card number has 6 digits (between 1 to 9). The firs [#permalink]
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16 Nov 2016, 10:54
mvictor wrote: I don't agree with the OA. 0 is divisible by 3, thus we can have: 7,8,9 in the 3rd digit 0,3,6,9  in the 4th digit the 6th digit can be 1,2,3 the 5th digit can be 3,6,9
total, we have 1x1x3x4x3x3 options.
we can eliminate D and E because both numbers are not divisible by 3. Please read the first part of the question carefully. Only digits from 19 are part of the credit card number sequence.
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Re: A credit card number has 6 digits (between 1 to 9). The firs [#permalink]
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09 Apr 2017, 17:10
Almost got fooled by this question but treated digits 5 & 6 as 1 combination
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Re: A credit card number has 6 digits (between 1 to 9). The firs [#permalink]
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09 Jul 2017, 03:32
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



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Re: A credit card number has 6 digits (between 1 to 9). The firs [#permalink]
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09 Jul 2017, 03:52
longhaul123 wrote: 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 31 62 93 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!
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Re: A credit card number has 6 digits (between 1 to 9). The firs [#permalink]
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03 Feb 2018, 22:58
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.
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Re: A credit card number has 6 digits (between 1 to 9). The firs [#permalink]
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03 Feb 2018, 23:14
Superg8 wrote: 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 Superg8Read 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!
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