Locked 627 in a aluminum performance center case

[colt22man]

Bought a 627, complete with performance center aluminum locking case. Problem is the case is locked and my FFL can't open it to record the revolver for transfer. The seller doesn't remember the combination. The default combination, 000 doesn't work. Any suggestions? I don't want to destroy the case.

 

[pharmer]

The good news is there are only 1000 permutations. The bad news is that you will need to get started, now. Joe

 

[Onomea]

Call the performance center and ask for ideas. Maybe they can put you in touch with the manufacturer of the case, and maybe they have ideas.

 

[3dots]

Locksmith?

 

[zzzippper]

I had to pop mine with a screw driver. Forget about locking it again.

 

[Arik]

Is that like that little briefcase looking thing? Take a dremel to the latches

 

[Alpo]

Oh hell. Pharmer has the right idea.

Sam's Club, maybe 30 years ago. They had a real nice leather briefcase on the clearance rack, because someone had changed the combination of both latches, and then locked it.

50 dollar case for 20 bucks. I bought it, took it to the house, sat down and started. 001, 002, 003, etc.

Took me ten minutes to get the first one and five for the second one. Sam's, basically, paid me 30 bucks for fifteen minutes work. I thought that was pretty good.

Doing mechanical damage to the lock and/or case is a last resort - if you try all 1000 combos - 000 to 999 - and it still does not open. Then you've got a bad latch.

 

[Phauxtoe]

Sit down and cycle thru the numbers, i had to do it once. it took about an hour. if the case is important to you.

OR
Look on YouTube to open it!
lots of ways like this...
[ame]https://www.youtube.com/watch?v=UhNvC38C2FU[/ame]

 

[ditrina]

Jus' Thinkin'

If order mattered, we would say select the first number (10 choices), then the second (9 choices) then the third (8 choices). So there would be 10 x 9 x 8 = 720 possible choices.
But because order does not matter, we have to take care of duplicates as you mentioned. How many duplicates are there for each set of three numbers? Well, again, we can choose one of the three as the "first", so there are 3 choices for that, then 2 choices for the "second" digit, then 1 choice for the last digit. There are, you see, 3 x 2 x 1 = 6 possible ways of arranging the three digits.
Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times. So we just divide by 6.

720 / 6 = 120.
​

That's your answer.
These are what are mathematically called "combinations". You can use a formula involving factorials to determine the number of combinations.
In this case, we say this is "10 Choose 3" and write it as 10C3. That means from a set of ten (digits in
this case), choose 3 regardless of order.
The formula for nCm is

nCm = n! / [ m! x (n-m)!]
​

So in your question, we have

10C3 = 10! / [3! x (10-3)!] = 10! / [3! 7!]
= (10 x 9 x 8 x 7!) / [ (3 x 2 x 1) 7!]
= (720) / 6 ... because the 7! on the top and bottom cancels.
= 120 as we expected.
​

 

[BigBill]

Sit down and cycle thru the numbers, i had to do it once. it took about an hour. if the case is important to you.

OR
Look on YouTube to open it!
lots of ways like this...
https://www.youtube.com/watch?v=UhNvC38C2FU

My old man did my lock like this it was 4265 she opened.

 

[BigBill]

Three digits is 999 start counting. Did u try 627? Or 357? My gut guess is 666, 696, 396, 693, 636, 639, but 666 is my first choice.

 

[Qball]

Try 357

 

[WPWarhawk]

Maybe try 627 ?

 

[Rustyt1953]

Hey ditrina:

 

[b_san]

Jus' Thinkin'

If order mattered, we would say select the first number (10 choices), then the second (9 choices) then the third (8 choices). So there would be 10 x 9 x 8 = 720 possible choices.
But because order does not matter, we have to take care of duplicates as you mentioned. How many duplicates are there for each set of three numbers? Well, again, we can choose one of the three as the "first", so there are 3 choices for that, then 2 choices for the "second" digit, then 1 choice for the last digit. There are, you see, 3 x 2 x 1 = 6 possible ways of arranging the three digits.
Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times. So we just divide by 6.

720 / 6 = 120.
​

That's your answer.
These are what are mathematically called "combinations". You can use a formula involving factorials to determine the number of combinations.
In this case, we say this is "10 Choose 3" and write it as 10C3. That means from a set of ten (digits in
this case), choose 3 regardless of order.
The formula for nCm is

nCm = n! / [ m! x (n-m)!]
​

So in your question, we have

10C3 = 10! / [3! x (10-3)!] = 10! / [3! 7!]
= (10 x 9 x 8 x 7!) / [ (3 x 2 x 1) 7!]
= (720) / 6 ... because the 7! on the top and bottom cancels.
= 120 as we expected.
​

The numbers can repeat and the placement of each number matters so there's a full 1000 combinations, nice math though even if it's not relevant.

 

[finesse_r]

For what those cases sell for it is well worth the effort to spin the dials a few hundred times to keep from having to damage the case.

PS Once you figure it out, write it down some place where you will not forget where you put it, but don't tape it to the case. LOL

Good luck with it. I have not had to crack combinations to locks since I was in junior high school. I was pretty good at picking locks and cracking combination locks back then, when my ears still worked well. I used to have to help people who forget the combination to their locks on their school lockers. LOL

 

[finesse_r]

Jus' Thinkin'

If order mattered, we would say select the first number (10 choices), then the second (9 choices) then the third (8 choices). So there would be 10 x 9 x 8 = 720 possible choices.
But because order does not matter, we have to take care of duplicates as you mentioned. How many duplicates are there for each set of three numbers? Well, again, we can choose one of the three as the "first", so there are 3 choices for that, then 2 choices for the "second" digit, then 1 choice for the last digit. There are, you see, 3 x 2 x 1 = 6 possible ways of arranging the three digits.
Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times. So we just divide by 6.

720 / 6 = 120.
​

That's your answer.
These are what are mathematically called "combinations". You can use a formula involving factorials to determine the number of combinations.
In this case, we say this is "10 Choose 3" and write it as 10C3. That means from a set of ten (digits in
this case), choose 3 regardless of order.
The formula for nCm is

nCm = n! / [ m! x (n-m)!]
​

So in your question, we have

10C3 = 10! / [3! x (10-3)!] = 10! / [3! 7!]
= (10 x 9 x 8 x 7!) / [ (3 x 2 x 1) 7!]
= (720) / 6 ... because the 7! on the top and bottom cancels.
= 120 as we expected.
​

This must be some of that there new math I keep hearing about. Lots of pretty numbers and symbols, but somehow I suspect that 120 is not the answer in the back of the math book.

 

[PatriotX]

Mine suffered the fate of the screwdriver, before it came into my possession.

Personally, I would have spun the dials.

 

[ditrina]

This must be some of that there new math I keep hearing about. Lots of pretty numbers and symbols, but somehow I suspect that 120 is not the answer in the back of the math book.

SPOILSPORT!!

 

[Exiled Cheesehead]

357 753

My 627-3 case has 357 for the left hand combo and 753 for the right hand combo.