18 Combo

break this code to get 10 points?

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). Proof of minimality not required.
hi some1 replied 40.do u have explanation for that.
i though its 50

40

edit
i'm sorry. my answer is wrong. your answer of 50 must be the right one. first, try all triple digits (000 to 999). those are my first 10 trials. then, i'll choose 40 other numbers carefully so that any one of them could represent all the numbers without repeating digits.

000 = representing 28 different combinations.
111
222
333
444
555
666
777
888
999
all 10 of the above covered 10*28 = 280, all with repetition within.

let's take an example : 123. it could represent :
120, 124, 125, 126, 127, 128, 129,
103, 143, 153, 163, 173, 183, 193,
023, 423, 523, 623, 723, 823, 923,
and of course, 123 itself. a total of 22 combos.

but then, with another number 185, it could represent 125 and 183, too. so there are double counting / overlapping involved. after considering the shared part, MAYBE 1 number is able to represent 18 combos by itself. so possible answer
= 10 + (10^3 - 28*10) / 18
= 10 + 720 / 18
= 50

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