Okay, here's my math problem... the keypad on my truck has 5 buttons, and a 5 digit code. This leads me to believe that there are 3125 possible codes, requiring 15625 button presses to try all of them.

However, upon testing, if i enter in the first 2 digits, the wrong third digit, then start over with the complete right code, i've pushed 8 digits, and opened the door.

how many button pushes at MAX would it take to open my door, if only the last 5 digits pushed matter?

anyone have any super duty permutation or factorial software to figure this out?

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quote:

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Originally posted by Dwaggy
Thats alot of words for someone that sells DVDs out of the back of a truck

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You owe me $20

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I'm aware, refer to THIS THREAD for details, but i'll be paypaling you later...

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quote:

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Originally posted by Dwaggy
Thats alot of words for someone that sells DVDs out of the back of a truck

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Well it sounds like you already have your answer. I'm pretty sure your max is 15625, code #3125. Your best is code #1 and 5 key presses.

Can't use factorials as you can reuse the same number.

But better yet why not find the factory code printed on the module?

::Let me elaborate with less numbers. Say it is 2 digits with a 0-9 scale

[digit 1] [digit 2] = 10 choices * 10 = 100 numbers or 200 key presses.

If you enter the code just right it will be the max. Think of the code as 99 and it is last on the list and you went down the list as 81..82..83 where each false number would reset the code. You wouldn't know it so you kept going. Lets assume you don't put 89 in before 90 and get lucky and find the code. If we are assuming MAX we are looking for the worst case scenario. It would take it 200 key presses to get it.::

Last edited by case sensitive on 09-07-2008 at 10:26 AM

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*waits 200 tries*

fingers crossed for ya

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quote:

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Originally posted by yoshy
*waits 200 tries*

fingers crossed for ya

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this thread gave me a sore head

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