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YouApostropheRe
#144900282Saturday, August 30, 2014 6:39 PM GMT

if there is a door that requires a 4 digit number passcode how many different possible passcodes are there
shalomimshelley
#144900359Saturday, August 30, 2014 6:40 PM GMT

only 1 passcode is possible - the correct one
Kaidou
#144900422Saturday, August 30, 2014 6:41 PM GMT

^
YouApostropheRe
#144900474Saturday, August 30, 2014 6:41 PM GMT

nuh uh nothing is certain until observed there are
YouApostropheRe
#144900520Saturday, August 30, 2014 6:42 PM GMT

oops i got cut off there are equally valid possibilities
neongoo51
#144900556Saturday, August 30, 2014 6:42 PM GMT

Indefinite. Because you didn't say if there were 2, 8, 10, or 16 different numbers to choose from. --[[ Shoot, I accidentally made another post.
miles94
#144900604Saturday, August 30, 2014 6:43 PM GMT

we have a winner
shalomimshelley
#144900605Saturday, August 30, 2014 6:43 PM GMT

no, there is only ONE passcode, you said so yourself "a 4 digit number passcode" "a" is singular in this case, since "passcode" is singular, you didn't say anything like "set of passcodes" therefore, only one passcode is needed
Negativize
#144900620Saturday, August 30, 2014 6:43 PM GMT

none smash the passcode social life is for stupid face
YouApostropheRe
#144900678Saturday, August 30, 2014 6:44 PM GMT

i said number passcode if not specified, then that covers the scope of all numbers
shalomimshelley
#144900728Saturday, August 30, 2014 6:44 PM GMT

"Because you didn't say if there were 2, 8, 10, or 16 different numbers to choose from" 4 digits = 4 single numbers only, from 0 to 9 "13" is TWO digits, not 1 digit there are only 10 (1,2,3,4,5,6,7,8,9,0) different kinds of single-digit numbers
neongoo51
#144900857Saturday, August 30, 2014 6:45 PM GMT

I was referring to binary, octal, decimal and hexadecimal counting systems. In hexadecimals, 13 = D. --[[ Shoot, I accidentally made another post.
YouApostropheRe
#144900888Saturday, August 30, 2014 6:46 PM GMT

everyone got it wrong hehehe
Negativize
#144900909Saturday, August 30, 2014 6:46 PM GMT

damnit guys just smash the fking door social life is for stupid face
neongoo51
#144901004Saturday, August 30, 2014 6:47 PM GMT

But either way, 2^4, 8^4, 10^4, and 16^4 for respective counting systems. I'm incorrect, right? --[[ Shoot, I accidentally made another post.
funnygirl686
#144901126Saturday, August 30, 2014 6:48 PM GMT

Around 9000 I think.
NottheNameIWant
#144901279Saturday, August 30, 2014 6:50 PM GMT

equation satchel charge + door + detonator = entry
funnygirl686
#144901297Saturday, August 30, 2014 6:50 PM GMT

Well OP, am I right?
TheCriminalTurkey
#144901336Saturday, August 30, 2014 6:50 PM GMT

10,000 What's this madness ye spew? I rate it 14/14 on the pH scale.
YouApostropheRe
#144901358Saturday, August 30, 2014 6:50 PM GMT

rong again!!111
obeyswaginsidejob
#144901459Saturday, August 30, 2014 6:51 PM GMT

0000 - 9999 total of 10999 hatflip
Lord_Zephyrus
#144901553Saturday, August 30, 2014 6:52 PM GMT

Technically 10^4 different possible combinations. In simpler terms; ten to the power of 4.
neongoo51
#144901637Saturday, August 30, 2014 6:53 PM GMT

No, this is some sort of trick question. Hmm. --[[ Shoot, I accidentally made another post.
neongoo51
#144901856Saturday, August 30, 2014 6:56 PM GMT

Fine, I'm just going to throw a wild guess and say ∞ possible combinations... --[[ Shoot, I accidentally made another post.
NottheNameIWant
#144901968Saturday, August 30, 2014 6:57 PM GMT

satchel charge + door + detonator = entry [2] floodcheck don't you dare glitch up again

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