Lets see how smart Sup Forums is
How many odd 3-digit whole numbers do not contain the digits 0 or 1?
Lets see how smart Sup Forums is
How many odd 3-digit whole numbers do not contain the digits 0 or 1?
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>There are 899 3 digit #s.
>There is one nore that is even than odds leaving 449 odds.
>Off those, 1/5 end in 1 and all #s that end in 0 are even.
The total left over is: 359
Incorrect
there are 32 of them in the 200's
so does that mean there are 32*8 of them?
idk what that is I only calculated this using my fingers
513
4*4*4, 0 is even, there are 5 odd numbers, excluding '1' we have only 4...
Incorrect
36/90 = 2/5 = .40 = 40%?
wtf kind of math are you doing?
90...
There are 90 odd 3 digit #s that do not end in 1 or 0
nevermind
do not contain != do not end...
only correct answer
256
The issue is you have to consider removing numbers like 118 because that has a 1 in it. so you have to remove the entire 100s list
332
256
The answer is 512, you dickless shithorse.
Is this it? I know its permutations and combinations but i dont remember anything from that part of precalc
Explain someone
256
Negative numbers be whole too, muthafucka
see
you can just count them in your head....
Easy peasy. The answer is x, where x represents the number of odd 3-digit whole numbers that do not contain the digits 0 or 1.
/thread
202
64.
there are 10 numbers: 0,1,2,3,4,5,6,7,8,9
even numbers are: 0,2,4,6,8
odd are: 1,3,5,7,9
we have five odd numbers
we do not want '1' in our odd numbers
it is a permutation, P(a) = a^x
a = four odd numbers, x = three because we want only 3-digit numbers... so 4^3=64
You guys are fucking retarded.
odd => ends in 1 3 5 7 or 9
excluding one, there are 4 choices.
the two other digits have 8 choices each.
8*8*4= 256
(function(){
a = 0;
for (i = 100; i < 1000; i++) {
s = "" + i;
if (
i % 2 == 1 &&
s.indexOf("0") == -1 &&
s.indexOf("1") == -1
) a++;
}
return a;
})();
256
you're right man
You be the retarded one, yo:
See
So the right answer is:
Dubs don't lie nigga
>255?
Congrats your the only non retard
Good job
Accidental greentext.
This is why the Chinese are better than you, dumbass.
Dubs layin' truth on all y'all muthafuckas
256 - just excluded everything manually.
(899-100-80-8*9-8*9-8*8+1)/2
There are 8 options for each digit.
It's 8 cubed
Good point.
...
Shit son
oh snap looks like someone else is a dickless shithorse
...
Don't care, cute girl.
check other sources, yo
Wolfram says it can go either way. Looks like OPs question might not have one clear answer.
Nope
thefreedictionary.com
some definitions exclude negatives, others don't. OP is a faggot for not stating the problem better.
OP is a faggot.
There's one thing we can all agree on.
Alright, problem solved.
From discrete mathematics:
8 choose 1
X
8 choose 1
X
8 choose 1
8 x 8 x 8 = 512