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Chris gave Brian a fourdigit code to disarm Chris’s security system.
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03 Dec 2019, 23:50
Question Stats:
46% (01:29) correct 54% (01:49) wrong based on 147 sessions
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Chris gave Brian a fourdigit code to disarm Chris’s security system. However, when Brian typed in the code, it didn’t work. If Brian realizes that Chris must have written exactly one digit incorrectly, but Brian does not know which digit is incorrect or what its correct value is, how many more attempts will Brian need to guarantee that he identifies the correct fourdigit code? A. 36 B. 39 C. 3,024 D. 6,561 E. 9,999 Are You Up For the Challenge: 700 Level Questions
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Chris gave Brian a fourdigit code to disarm Chris’s security system.
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Updated on: 08 Dec 2019, 20:05
Bunuel wrote: Chris gave Brian a fourdigit code to disarm Chris’s security system. However, when Brian typed in the code, it didn’t work. If Brian realizes that Chris must have written exactly one digit incorrectly, but Brian does not know which digit is incorrect or what its correct value is, how many more attempts will Brian need to guarantee that he identifies the correct fourdigit code? A. 36 B. 39 C. 3,024 D. 6,561 E. 9,999 Are You Up For the Challenge: 700 Level QuestionsExplanation: _ _ _ _ As 0 to 9 can be fitted for above 4 numbers. Also, we don't know which number is incorrect. He had check every number from 0000 to 9999, minus 1 ( as one combination will be correct) Total Tries needed = 10 X 10 X 10 X 10  1 = 9999 CCCW can be arranged in 4 ways W can be corrected in 9 ways Total number of ways = 4 * 9 = 36 IMOA



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Re: Chris gave Brian a fourdigit code to disarm Chris’s security system.
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08 Dec 2019, 08:15
for a 4 digit code 1 is wrong so total ways to match code ; CCCW ; 4!/3! ; 4 ways of combination is possible now since 09 are 10 digits and the one W is not correct so we have 9 digits left total possible ways ; 4*9 ; 36 IMO A Bunuel wrote: Chris gave Brian a fourdigit code to disarm Chris’s security system. However, when Brian typed in the code, it didn’t work. If Brian realizes that Chris must have written exactly one digit incorrectly, but Brian does not know which digit is incorrect or what its correct value is, how many more attempts will Brian need to guarantee that he identifies the correct fourdigit code? A. 36 B. 39 C. 3,024 D. 6,561 E. 9,999 Are You Up For the Challenge: 700 Level Questions



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Re: Chris gave Brian a fourdigit code to disarm Chris’s security system.
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11 Dec 2019, 09:41
aarushisingla wrote: Hi,
How can 0 be in the thousands place because if it would be , then it will make 3digit code. I guess it should be 9*3 (because there are 9 numbers and since 0 cannot taken thousands place it will not be counted.) Hence, for thousands place, there should be 8 numbers (excluding 0)
So, Total ways = 27 + 8 = 35.
Please tell me if I am wrong. Hi aarushisingla , 09 are 10 digits and the one W is not correct so we have 9 digits left Since In Lock , 0 can be also put. Lock Password can be 0000 or 0900 or 0090 or 0000. Hope U got my Point. Please Do PM , if You didn't get it. Regards, Rajat Chopra



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Re: Chris gave Brian a fourdigit code to disarm Chris’s security system.
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12 Dec 2019, 18:57
Bunuel wrote: Chris gave Brian a fourdigit code to disarm Chris’s security system. However, when Brian typed in the code, it didn’t work. If Brian realizes that Chris must have written exactly one digit incorrectly, but Brian does not know which digit is incorrect or what its correct value is, how many more attempts will Brian need to guarantee that he identifies the correct fourdigit code? A. 36 B. 39 C. 3,024 D. 6,561 E. 9,999 Are You Up For the Challenge: 700 Level QuestionsSuppose that Brian typed in “0000”. From the information given to us, we understand that the correct code contains three digits of “0” and one nonzero digit. If the first digit is incorrect, the correct code could be any one of 1000, 2000, …, 9000 (nine codes). Similarly, if the second digit is wrong, the correct code could be any one of 0100, 0200, …, 0900 (nine codes). Following the same logic, we see that If the third or fourth digits are wrong, there will be 9 possibilities for the correct code in each case. Thus, to guarantee that Brian identifies the correct code, he needs to try at most 9 + 9 + 9 + 9 = 36 codes. Note: Technically, if he only needs to “identify” the correct code, he needs at most 35 attempts. If he does not get the correct code in the first 35 attempts, he can then be sure that the correct code is the only remaining code that he did not try. Answer: A
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Re: Chris gave Brian a fourdigit code to disarm Chris’s security system.
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22 Jan 2020, 15:59
altairahmad wrote: Please one basic confusion
Only one digit W (09) will be correct and only 1 arrangement will be correct from following 4 arrangements right ?
WCCC , CWCC, CCWC, CCCW
So doesn't it mean that he will have to try 10 x 3 (3 wrong arrangements with full 10 digits) + 9 (1 correct arrangement with 1 correct digit) = 39 attempts ?
Lets say the correct code is 0009. You are given 0000. Start from the beginning (left) and work your way to the right. You try 0000 initially (from the question) and that doesn't work. So you try nine more times: 1000, 2000, ...., 9000. Okay, they all didn't work, so try the 2nd digit now. You don't need to try 0000 again since you did that already. You try 0100, 0200, 0300, 0400, ...., 0900. None of those work, so try the 3rd digit. You don't need to try 0000 again since you did that already. You try 0010, 0020, 0030,....., 0090. None of those work so you try the 4th digit. You don't need to try 0000 again since you did that already. You try 0001, 0002, 0003, and then on the last try (you're very unlucky!) 0009 works! So 36 additional tries were needed.



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Re: Chris gave Brian a fourdigit code to disarm Chris’s security system.
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28 Jan 2020, 06:31
altairahmad wrote: ScottTargetTestPrep Bunuel KinshookPlease one basic confusion Only one digit W (09) will be correct and only 1 arrangement will be correct from following 4 arrangements right ? WCCC , CWCC, CCWC, CCCW So doesn't it mean that he will have to try 10 x 3 (3 wrong arrangements with full 10 digits) + 9 (1 correct arrangement with 1 correct digit) = 39 attempts ? Your reasoning is almost correct, just notice that there are not 10 but 9 possibilities for W in each case. It is given that Brian tries a code which does not work. For each of the arrangements WCCC, CWCC, CCWC and CCCW; Brian does not need to consider one digit because of that one code he tried in the beginning. For instance, suppose Brian tried 0000 in the beginning and the actual code was 0009. For the arrangement WCCC, Brian only needs to test the values 19 for W. For the arrangement CWCC, again he needs to only test 19, etc. Thus, if you decrease the number of options for W by 1 and do the same calculation again, you will get the correct answer of 9 + 9 + 9 + 9 = 36 more tries.
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Re: Chris gave Brian a fourdigit code to disarm Chris’s security system.
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08 Dec 2019, 08:32
Bunuel wrote: Chris gave Brian a fourdigit code to disarm Chris’s security system. However, when Brian typed in the code, it didn’t work. If Brian realizes that Chris must have written exactly one digit incorrectly, but Brian does not know which digit is incorrect or what its correct value is, how many more attempts will Brian need to guarantee that he identifies the correct fourdigit code? A. 36 B. 39 C. 3,024 D. 6,561 E. 9,999 Are You Up For the Challenge: 700 Level QuestionsTotal number of ways = 4 * 9 = 36 since CCCW can be arranged in 4 ways and W can be corrected in 9 ways. IMO A



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Re: Chris gave Brian a fourdigit code to disarm Chris’s security system.
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11 Dec 2019, 06:02
Hi,
How can 0 be in the thousands place because if it would be , then it will make 3digit code. I guess it should be 9*3 (because there are 9 numbers and since 0 cannot taken thousands place it will not be counted.) Hence, for thousands place, there should be 8 numbers (excluding 0)
So, Total ways = 27 + 8 = 35.
Please tell me if I am wrong.



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Chris gave Brian a fourdigit code to disarm Chris’s security system.
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22 Jan 2020, 00:07
ScottTargetTestPrep Bunuel KinshookPlease one basic confusion Only one digit W (09) will be correct and only 1 arrangement will be correct from following 4 arrangements right ? WCCC , CWCC, CCWC, CCCW So doesn't it mean that he will have to try 10 x 3 (3 wrong arrangements with full 10 digits) + 9 (1 correct arrangement with 1 correct digit) = 39 attempts ?



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Re: Chris gave Brian a fourdigit code to disarm Chris’s security system.
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23 Jan 2020, 10:41
kendannedy wrote: altairahmad wrote: Please one basic confusion
Only one digit W (09) will be correct and only 1 arrangement will be correct from following 4 arrangements right ?
WCCC , CWCC, CCWC, CCCW
So doesn't it mean that he will have to try 10 x 3 (3 wrong arrangements with full 10 digits) + 9 (1 correct arrangement with 1 correct digit) = 39 attempts ?
Lets say the correct code is 0009. You are given 0000. Start from the beginning (left) and work your way to the right. You try 0000 initially (from the question) and that doesn't work. So you try nine more times: 1000, 2000, ...., 9000. Okay, they all didn't work, so try the 2nd digit now. You don't need to try 0000 again since you did that already. You try 0100, 0200, 0300, 0400, ...., 0900. None of those work, so try the 3rd digit. You don't need to try 0000 again since you did that already. You try 0010, 0020, 0030,....., 0090. None of those work so you try the 4th digit. You don't need to try 0000 again since you did that already. You try 0001, 0002, 0003, and then on the last try (you're very unlucky!) 0009 works! So 36 additional tries were needed. Nicely explained. Thanks a bunch.




Re: Chris gave Brian a fourdigit code to disarm Chris’s security system.
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