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Chris gave Brian a four-digit code to disarm Chriss security system. Updated on: 08 Sep 2025, 23:37
Originally posted by Bunuel on 04 Dec 2019, 00:50.
Last edited by Bunuel on 08 Sep 2025, 23:37, edited 1 time in total.
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47% (01:38) correct 53% (01:47) wrong based on 606 sessions

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Chris gave Brian a four-digit 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 types in the correct four-digit code?

A. 36
B. 39
C. 3,024
D. 6,561
E. 9,999


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A
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Chris gave Brian a four-digit code to disarm Chriss security system. Updated on: 08 Dec 2019, 21:05
Originally posted by rajatchopra1994 on 04 Dec 2019, 01:52.
Last edited by rajatchopra1994 on 08 Dec 2019, 21:05, edited 1 time in total.
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Bunuel
Chris gave Brian a four-digit 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 four-digit 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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Explanation:
_ _ _ _
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

IMO-A
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Chris gave Brian a four-digit code to disarm Chriss security system. 08 Dec 2019, 09:15
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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 0-9 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
Chris gave Brian a four-digit 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 four-digit 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 four-digit code to disarm Chriss security system. 08 Dec 2019, 09:32
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Bunuel
Chris gave Brian a four-digit 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 four-digit 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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Total 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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Chris gave Brian a four-digit code to disarm Chriss security system. 11 Dec 2019, 07:02
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Hi,

How can 0 be in the thousands place because if it would be , then it will make 3-digit 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 four-digit code to disarm Chriss security system. 11 Dec 2019, 10:41
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aarushisingla
Hi,

How can 0 be in the thousands place because if it would be , then it will make 3-digit 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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Hi aarushisingla ,

0-9 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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Chris gave Brian a four-digit code to disarm Chriss security system. 12 Dec 2019, 19:57
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Bunuel
Chris gave Brian a four-digit 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 four-digit 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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Suppose that Brian typed in “0000”. From the information given to us, we understand that the correct code contains three digits of “0” and one non-zero 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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Chris gave Brian a four-digit code to disarm Chriss security system. 22 Jan 2020, 01:07
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ScottTargetTestPrep Bunuel Kinshook

Please one basic confusion

Only one digit W (0-9) 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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Chris gave Brian a four-digit code to disarm Chriss security system. 22 Jan 2020, 16:59
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altairahmad


Please one basic confusion

Only one digit W (0-9) 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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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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Chris gave Brian a four-digit code to disarm Chriss security system. 28 Jan 2020, 07:31
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altairahmad
ScottTargetTestPrep Bunuel Kinshook

Please one basic confusion

Only one digit W (0-9) 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 ?
Show more

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 1-9 for W. For the arrangement CWCC, again he needs to only test 1-9, 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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Chris gave Brian a four-digit code to disarm Chriss security system. 03 Aug 2021, 04:33
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Bunuel
Chris gave Brian a four-digit 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 four-digit 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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Let the number entered be 0001, which is a wrong code.

Worst case scenario is that the code is 0000, and he checks the digits from left to right and in ascending order.

The first will be 1001 to 9001.......9 ways.
Next will be 0101 for the second digit till 0901....9 ways
Similarly 9 ways for third digit, and last number checked will be 0091.
Next for the 4th digit, we will start from 0002 and go upto 0009......8 ways

If we are still not able to open it, the correct code has to be 0000, and we do not require to test it.
Thus, we require 35 MORE attempts to IDENTIFY the correct code.

The question is wrongly worded. It should have been: how many more attempts will Brian need to guarantee that he opens the security system.
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Chris gave Brian a four-digit code to disarm Chriss security system. 06 Aug 2021, 22:09
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Could someone tell whether this is an acceptable solution:
->There are 40 ways in which he can try a different code.
->4 of those ways would be repeated.
Final Answer: 40-4=36 ways.
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Chris gave Brian a four-digit code to disarm Chriss security system. 16 Nov 2022, 21:38
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Bunuel
Chris gave Brian a four-digit 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 four-digit code?

A. 36
B. 39
C. 3,024
D. 6,561
E. 9,999


Are You Up For the Challenge: 700 Level Questions

Suppose that Brian typed in “0000”. From the information given to us, we understand that the correct code contains three digits of “0” and one non-zero 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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In agreement.

When he enters the 35th attempt, at that point he would know that the next code he types in will work. Therefore, he shouldn’t have to type it in.

The correct code has been “identified.”

Be that as it may, the answer can not be more than 36, so A.

Posted from my mobile device
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Chris gave Brian a four-digit code to disarm Chriss security system. 08 Sep 2025, 18:25
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You caught me! We later fixed the wording, but unfortunately GMAT Club is forever, and I have no way to edit the wording in this problem. But you're totally right; it should say "to guarantee that he types in the correct four-digit code." Kudos!


chetan2u


Let the number entered be 0001, which is a wrong code.

Worst case scenario is that the code is 0000, and he checks the digits from left to right and in ascending order.

The first will be 1001 to 9001.......9 ways.
Next will be 0101 for the second digit till 0901....9 ways
Similarly 9 ways for third digit, and last number checked will be 0091.
Next for the 4th digit, we will start from 0002 and go upto 0009......8 ways

If we are still not able to open it, the correct code has to be 0000, and we do not require to test it.
Thus, we require 35 MORE attempts to IDENTIFY the correct code.

The question is wrongly worded. It should have been: how many more attempts will Brian need to guarantee that he opens the security system.
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Chris gave Brian a four-digit code to disarm Chriss security system. 08 Sep 2025, 22:57
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since CCCW can be arranged in 4 ways and W can be corrected in 9 ways.
Total number of ways = 4 * 9 = 36
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Chris gave Brian a four-digit code to disarm Chriss security system. 08 Sep 2025, 23:37
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AnthonyRitz
You caught me! We later fixed the wording, but unfortunately GMAT Club is forever, and I have no way to edit the wording in this problem. But you're totally right; it should say "to guarantee that he types in the correct four-digit code." Kudos!



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Fixed the issue. Thank you!
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Chris gave Brian a four-digit code to disarm Chriss security system. 09 Sep 2025, 02:12
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AnthonyRitz
You caught me! We later fixed the wording, but unfortunately GMAT Club is forever, and I have no way to edit the wording in this problem. But you're totally right; it should say "to guarantee that he types in the correct four-digit code." Kudos!



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Thanks Anthony 😊
Bunuel seems to have already fixed it.
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Chris gave Brian a four-digit code to disarm Chriss security system. 17 Nov 2025, 23:18
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Hi KarishmaB MartyMurray chetan2u

Is the below logic wrong because we are given that only one digit is entered incorrectly? Please let me know if any other reason why the below answer is wrong.
Total Tries needed = 10 X 10 X 10 X 10 - 1 = 9999
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Chris gave Brian a four-digit code to disarm Chriss security system. 18 Nov 2025, 02:57
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Yes, this assumes that any other 4 digit code is possible (except the one he has in his hand)
Say he has 2314.
Your logic assumes that 1785 is possible. But is it? No. Only 1 digit is wrong.

agrasan
Hi KarishmaB MartyMurray chetan2u

Is the below logic wrong because we are given that only one digit is entered incorrectly? Please let me know if any other reason why the below answer is wrong.
Total Tries needed = 10 X 10 X 10 X 10 - 1 = 9999
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