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  • Break the security

    arsh86

    Member

    Updated: Oct 25, 2024
    Views: 1.0K
    You want to get through a security door, where you have to enter 3 digits as password, where each digit can be any from 0 to 9. The checking mechanism is, however, defective, and so will let you in if any 2 of your digits match with the password's. e.g. if the password is 087 and you enter 057, then you will be let in. You don't have much time and hence want to make as few tries as possible. What is the minimum number of tries in which you can definitely enter, and what are they (you need not give explicitly, if you describe a pattern).



    ...give your answer with proof.
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Replies

  • raj87verma88

    Member • Mar 21, 2009

    Why do you wish to break through a "Security Door"?
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  • Kaustubh Katdare

    Administrator • Mar 21, 2009

    I knew a method of solving this type of problems. Will have to scratch my skull for this one. Hmm! :neutral:
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  • ms_cs

    Member • Mar 21, 2009

    Let me know the answer?
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  • Yamini L

    Member • Mar 22, 2009

    I think answer for this should be 50
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  • Saandeep Sreerambatla

    Member • Mar 22, 2009

    Have got two methods of solving and got answers as 50 once and 55 for the next method.

    So waiting what the person asked will tell ..😎😎
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  • arsh86

    Member • Mar 23, 2009

    English-Scared
    Have got two methods of solving and got answers as 50 once and 55 for the next method.

    So waiting what the person asked will tell ..😎😎

    can u explain both methods???????????
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  • Differential

    Member • Mar 24, 2009

    Is the digit matching positional?
    I mean, if pasword is 087, then it won't accept 507. Tough 2 digsits are there, they are at diff positions. RIGHT?
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  • Saandeep Sreerambatla

    Member • Mar 24, 2009

    Differential
    Is the digit matching positional?
    I mean, if pasword is 087, then it won't accept 507. Tough 2 digsits are there, they are at diff positions. RIGHT?

    Correct.
    pwd=087 if you give 057 it is correct.
    😎
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  • Raviteja.g

    Member • Mar 24, 2009

    yes, if it does not care about positions then the number of tries would be decreased to great extent.
    otherwise it would be critical .
    anyway waiting for the answer........
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  • ms_cs

    Member • Mar 25, 2009

    ... I like these type of problems...I need to see these type of ...Send the link here or send the book name
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  • Differential

    Member • Mar 25, 2009

    I tried to solve this one and got the answer as 91. Am I correct or what?

    😕
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  • Raviteja.g

    Member • Mar 25, 2009

    Differential,
    can you give explanation
    then we can analyze whether you are correct or went wrong anywhere
    ok
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  • Raviteja.g

    Member • Mar 30, 2009

    I tried to solve this one
    where is the explanation boss!
    I think you guessed the answer .
    Am I right Differential!!!?
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  • Differential

    Member • Mar 30, 2009

    Raviteja.g
    where is the explanation boss!
    I think you guessed the answer .
    Am I right Differential!!!?
    I have an explanation! I'm waiting to hear if my answer is right or wrong.
    There is no point in telling the explanation if answer is wrong.
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  • sarveshgupta

    Member • Mar 31, 2009

    is the answer 100?
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  • Saandeep Sreerambatla

    Member • Apr 1, 2009

    The answer should be either 50 or 55
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  • Digs

    Member • Apr 30, 2009

    According to me the answe is 61...i found it out by using a simple mathematical technique..is it right??
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  • RajdeepCE

    Member • May 16, 2009

    The number is 99, if it concerns the position. Cause we have to guess 2 number & there are 99 possiblities.
    But if the position is not concerned than the number of tries wilk be decreased. So as per my calculation the number of tries for this method is 55. The explanation for is that for each series e.g. 0-9, 10-19, 20-29,..., 90-99. The combination will decrease one for each series, cause the number is repeated already. That means series will be 0-9, 11-19, 22-29, 33-39,...,99. So the total number of tries will be 55.
    Am I right? I am very keen to know the answer if I am wrong.
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  • Differential

    Member • May 17, 2009

    Looks like the one who asked the question and knows the real answer is away for long. Let's wait for him.
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