Opening Combo Padlocks

[LCSBSSRHXXX]

Opening Combo Padlocks:
By LCSBSSRHXXX

I learned this a while ago from a friend of mine its pretty cool, and easy to do, but it can take time...
This is the type of lock I am refering to when I say Combo Padlock, the more digits on the lock the longer it takes to open it.

Ok, to start out you set the lock to 0,0,0 now pull down on the lock so there is pressure so when you get the combo it will pop right open. Now twist the lowest wheel, cycle through it untill you get back to 0, than set the second dial to 1 and repeat, do this untill the second wheel has gotten back to 0 than turn the third wheel to 1 and go back to the first wheel, repeate this entire procedure until the lock opens.

So if the combo is 1,2,3 and your trying to figure it out you would do this:
Row 3:

0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000111111111111111111111111

Row 2:

0000000000111111111122222222223333333333444444444455555555556666666666777777777788888888889999999999000000000011111111112222

Row 1:

0123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123

[hba]

In Row 1, you're missing an 8 in the first ten digits.

And yes, this would take for f*cking ever just to figure it out.

You'll need more time to open it than what time it will take to still keep it convenient to why you were even trying to open the lock for in the first place.

And I believe you can figure out how many possible combinations there are by starting with your highest figure and multiplying by the next figure in descending order until 1.

9*8*7*6*5*4*3*2*1 = 362,880 different combinations?

[LCSBSSRHXXX]

It doesn't take long, you just spin through it.
Edit: Tanks for pointing out that little mistake, I fixed it.

the total combinations should be 10 x 10 x 10 = 1000. : D

[hba]

No...

You just cubed the amount of numbers on the combination dial...combinational outcomes are different from just how many numbers are the dial(s).

Let's take a simple imaginary lock with only 0, 1, 2, and 3 as the digits.

Now with your logic, you'd take 4^3 and think the number of combinations is 64, when there's really only 24.

[LCSBSSRHXXX]

4^3 = 64 not 12

[hba]

Yeah, I fixed. I thought * was ^ for some reason. :P

[DA(]

Actually your both wrong. Have you ever had the class called probability and statistics? There is a formula for combinations but I don't know how to post it here.

[hba]

This is the logic I used for figuring how many combinations there are. Except it's not combinations, its permutations.

A permutation of a set of objects is an arrangement of the objects in a certain order. For example, take the set of four objects {pepperoni, sausage, onions, mushrooms}. They can be arranged on a pizza many different ways. Below are a few of the ways.

pepperoni, sausage, onions, mushrooms
sausage, onions, mushrooms, pepperoni
onions, mushrooms, pepperoni, sausage
mushrooms, pepperoni, sausage, onions
pepperoni, sausage, mushrooms, onions

There are some more, but we won't list them. To find the number of different arrangements of the set we select a first choice; there are 4 possible choices. Now we take a second choice; there are 3 choices. Now pick a third choice; there are 2 choices. Finally, there is 1 choice for the last selection. Thus, there are 4 * 3 * 2 * 1 or 24 different ordered arrangements of the toppings. This product can also be written as 4! (read: 4-factorial).

http://library.thinkquest.org/20991/alg2/prob.html

~And on note, don't say we're both wrong and then not prove your point of us being wrong. It's disrespectful and makes you look like a jackass.

[DA(]

HBA got it, kind of.

[arpsmack]

Umm Akazukin was right...

HBA's answer is incorrect because the numbers composing a combination to a lock can be repeated. He is suggesting that if 8 is the first number in the combination, then it can't be the second or third, which is not true (i.e. 8-4-8 is a valid combination.)

For the first number of the combination you can use any value in the set 0-9. Same for the second number, and same for the third as well. Hence there are 10 x 10 x 10 combinations... or 1000.

You are at least correct when you say that a combination lock really isn't a combination lock; it should be called a permutation lock since the order of the digits matters (i.e. 1-9-8 is not the same as 9-1-8.)

[hba]

Originally Posted by arpsmack

Umm Akazukin was right...

HBA's answer is incorrect because the numbers composing a combination to a lock can be repeated. He is suggesting that if 8 is the first number in the combination, then it can't be the second or third, which is not true (i.e. 8-4-8 is a valid combination.)

For the first number of the combination you can use any value in the set 0-9. Same for the second number, and same for the third as well. Hence there are 10 x 10 x 10 combinations... or 1000.

You are at least correct when you say that a combination lock really isn't a combination lock; it should be called a permutation lock since the order of the digits matters (i.e. 1-9-8 is not the same as 9-1-8.)

Dude, if you didn't see my example...

pepperoni, sausage, onions, mushrooms
sausage, onions, mushrooms, pepperoni
onions, mushrooms, pepperoni, sausage
mushrooms, pepperoni, sausage, onions
pepperoni, sausage, mushrooms, onions

You can see that I'm not withdrawing the fact that a number can be used in the same slot more than once.

Thanks for the knowledge HBA, and others for comments. The real concern is about breaking the lock ASAP, right?

[Belphegor]

I remember doing this thing in math class... There was a statue, with 8 hands. Each hand held a diffrent object. So for all the combinations of them 8 hands holding a diffrent object, the number came out to like 12 digits long. T'was crazy.