my cheap push-button combination padlock..

[captainjerry]

Ever seen one of these before?

got it some time ago, for something like $5..
the package proudly proclaimed that this lock has got 40,000 possible combinations.

here are the facts:
1. first of all, it comes with one code from the factory (which can't be changed)
2. the code consists of 5 digits
3. the order of the buttons pushed is not important. meaning that if the code is 18405, 08415 (or any other number consisting of those digits..) will work just as well.
4. and no "half-way" push-buttons..



well, using Newton's binomial theorem, I found out that this lock only has 252 (!) possible combinations..




what do you all think?

of course, anyone can take half an hour and simply try all 252 possible combinations...
but any other ideas on how to pick this sort of lock?

[josh0094]

Great work! you might be able to shim that.

[Trip Doctor]

I've seen this exact same lock before, it was attached to a locker. I wanted to mess around with it, but I had no idea who's it was and the owner wasn't there.

I don't see how the Binomial Expansion would be applicable in trying to figure out the number of possible combinations here, and I'm pretty sure you did something wrong, as this lock has way more than 252 combinations.

[vrocco]

Can it be decoded using something akin to the simplex method?

[Eyes_Only]

I've seen this exact same lock before, it was attached to a locker. I wanted to mess around with it, but I had no idea who's it was and the owner wasn't there.

I don't see how the Binomial Expansion would be applicable in trying to figure out the number of possible combinations here, and I'm pretty sure you did something wrong, as this lock has way more than 252 combinations.

Yeah but he must have done something right cos he got it open.

[captainjerry]

I don't see how the Binomial Expansion would be applicable in trying to figure out the number of possible combinations here, and I'm pretty sure you did something wrong, as this lock has way more than 252 combinations.

could be.. to be hounest, math had always been my weakest class

this is how i did it..

n marking the number of buttons
so 'n=10'

and k marking the number of digits in the combination
therefore, 'k=5'

so..
= 252

as far as i know this formula can only be used with these sort of locks..

the conditions being that
1. each digit can only come out once in the combination (once pushed, the button can't be pushed again..)
2. the order of the digits doesn't matter (as I mentioned in my first post)..

for most Wheel combination locks this simple formula should reveal the number of possibilities

n = number of digits (that would be 10 in most cases )
k = number of wheels

[Trip Doctor]

Ah, I though you were doing the series expansion or something. (I also forgot about one of the conditions.)
One thing though.. that number, is a lot bigger than 252.
10!/5! =10(9)(8)(7)(6), and the calculator on my cumouter says that's 30240 (which is still less than 40,000).

[Trip Doctor]

Arighty, I see what you did. You had 5!*5!. I think the formula you would use in this case would be n!/(n-k)!, rather than the one you would use for the expansion with a k! on the bottom.
I never liked series o_O, they're right up there with linear algebra on my list of math subjects I hated doing (cool and useful nevertheless).

[captainjerry]

I never liked series o_O, they're right up there with linear algebra on my list of math subjects I hated doing.

well, at least we agree on one thing

you might be able to shim that.

thanks
can't see any use in trying to shim it though,
since there aren't any visible notches on the shackle..

i think i'll go get myself another one, and take it apart..
diagram maybe? someone?

[Eyes_Only]

LP101, one of the few sites on the web where chemists, metallurgists, mathematicians, locksmiths and hobbyists come together for a unified goal.

[Trip Doctor]

LP101, one of the few sites on the web where chemists, metallurgists, mathematicians, locksmiths and hobbyists come together for a unified goal.

LOL. There's a good campaign .

[Raymond]

Another way to say or think the same math formula in plain English is: How many choices do you have for the first possible number? -10. How many for the second? -9 Etc., etc. 10 x 9 x 8 x 7 x 6 = 30240. But, the combo can be entered in any order. Therefore, How many choices for the first number of the combo using the known correct numbers? -5. How many choices for the second number...-4 and so on. 5 x 4 x 3 x 2 x 1 = 120.
30240 / 120 = 252

Pick these simple locks by applying pressure to the shackle and push each button looking for the ones that are "tightened". Five will be very tight and the lock will open when you push the fifth button.

[vrocco]

Pick these simple locks by applying pressure to the shackle and push each button looking for the ones that are "tightened". Five will be very tight and the lock will open when you push the fifth button.

Yeah that's what I was talking about.

[greyman]

well, using Newton's binomial theorem, I found out that this lock only has 252 (!) possible combinations..

As far as I know, the binomial theorem is not due to Newton. The number of ways to choose 5 things from 10 is a high school maths problem: (10, 5)=10 C 5=252. No surprises there. Are you sure the manufacturer claims 40,000 combinations? If so, it may be possible to have other combinations that don't use 5 buttons or maybe order is important sometimes?

[Trip Doctor]

But, the combo can be entered in any order. Therefore, How many choices for the first number of the combo using the known correct numbers? -5.

Ah, yes, that's right Raymond. Sorry for confusing both of us Jerry .