How Long Would It Take to Try Every Combination on a 0-59 Dial Lock?

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The discussion revolves around calculating the time it would take to try every combination on a 0-59 dial lock, considering both three and four-number combinations. The total number of combinations is calculated as 60^3 for three numbers and 60^4 for four numbers, resulting in 13,176,000 possibilities. With each attempt taking 15 seconds, the total time required is approximately 2,287.5 days. There is debate about whether to account for the direction of turning the dial, which could potentially double the combinations. Ultimately, the conclusion is that the answer remains uncertain due to insufficient information about the lock's mechanics and tolerances.
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Help me solve this maths riddle!

John forgot to bring his house keys with him,so he must use secret numbers to open his house door.But he reasiled that he totally forgot the numbers,despite of that,he still insisted using the secret numbers. When he got there, he came to a combination lock on the door, with the dial numbers going from 0 to 59. Unfortunately, he forgot whether there were three or four numbers in the combination, or even which direction to turn the wheel!



If it takes him 15 seconds to try a single combination, how many days will it take him to to try every possible combination? Please round to the nearest day.


please help..i rephrase it a little...help please..i need the solution ASAP..as i need to pass up by tommorrow!
 
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From 0 to 59, that are 60 possibilities, I assume? (is that including or excluding 0 and/or 59?)
60^3 possibilities with 3 numbers.
60^4 possibilities with 4 numbers.
60^3+60^4=13176000 possibilities
13176000*15=197640000 seconds=3294000 minutes=54900 hours=2287.5 days.
 
I just saw exactly the same question on "math message board" except that it was phrased in terms of a bank robber trying to open a vault!
 
Should you multiply that answer by 2 because he doesn't know what direction to start turning and there's 2 possiblities there as well?
 
They should have told you exactly what type of lock this was. I assume koroljov's approach is correct for a type of lock that you would find on a http://www.hevanet.com/gillesp/briefcase-lock.JPG a student would put on their locker at school, which changes things. In that case, you would definitely have to mutiply both the number of combinations of a 3 number combination and a 4 number combination by 2 (for going to the right and then to left) and then add them, etc. But... that assumes for each combination that your dialing in a number, turning left, dialing in another, turning left, etc., and my experiences have always been dialing in a number and then turning the opposite direction of the previous turn, which would change the answer. I would also want to know whether or not to assume that combinations like 0-0-0, 1-1-1-1, 0-1-0 are valid entries.
 
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lol... gech. With your picture links, there's no guessing what it is you're talking about. Since direction was a concern, I imagined that they were talking about the high school locker kind of lock. But house locks are usually of the briefcase style. Assuming they're talking about a padlock, I remember on my high school locker lock that you didn't have to get it quite exact anyway. If your combination were 37-24-12 and you did 38-23-11, it would usually open. The numbers on the dial were a more accurate scale than the mechanics of the lock required. That changes the problem a bit too. So the real answer is "Unknown--not enough information given. Need to know the tolarance of the lock". Dohydj, teachers usually don't like these kind of answers.
 
try this

i assume i solve it unless their is misunderstanding to the given
(2 * 60)^4 * 15 / 3600 / 24 =36000 days =98.6 years

(as two right or wrong questions has two probabilities and they are two they have four combinations
3 right or wrong questions are 2^3 and so on......)
so one can solve number of possible combinations of 500 right or wrong questions is 10^(500log2) as 2^500 gives math error on calculator
 
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