Math help for the clincly insaine

[MLLMillenium]

So I will be. Very soon. My teacher argued there is nothing wrong with the question. So apparently.. this is the correct wording.. yet I doubt it becuase it makes no sense at all. Anyone out their not as dumb as me willing to help. +Scene points for anyone who can! lol.

Does a prime number, p, that is smaller than a number m, divide the number

N=(1*2*3*...*(m-1)*m)+1


(How is that Even an QUESTION! I dont get it at all!)

[burntheaction]

haha. sucks for you. i hate maths.

and no, it doesn't make sense.

[YourLatestVicti]

I completely confused and my friend who's fairly intelligent is just as confused.

[MLLMillenium]

I completely confused and my friend who's fairly intelligent is just as confused. So is like.. a whole dorm building. We have some time before it's due. Were all lost though. He said the question is written correctly. Maybe were all just stupid. Or maybe he is just a complete asshole, and likes to mess with us.

[YourLatestVicti]

The only answer my friend could get was "Any number can divide another number, unless it's 0." Or my personal favorite, "that doesn't make any sense."

[IcedOpethBlind]

Does a prime number, p, that is smaller than a number m, divide the number



yes, but im sure thats not all thats the answer.

[ghostyouare]

After you said "an question" I'm gonna say you are wrong.

[Ryan Mills]

Silly kids and their pointless math problems.

[ghostyouare]

Silly kids and their pointless math problems. Are you a trix kid?

[Ryan Mills]

Oh, and the answer is of course it does, how is that hard to see? (nope, it's not, see addendum).

A prime number is a number that has only itself and one as a divisor, like 7 or 13 or whatever. If you take a number greater than that number (say you take 15, which is greater than 13) and multiply it by every number underneath it until 1, you obviously multiply it by the number p (here 13). Thus the result is divisible by p (13) because the number was a product of p (13) and different numbers. Thus the answer is a whole number equal to m * (m-1) * (m-2) ..... * 1 excluding p itself. Hence p divides the product into some whole number because it was a part of creating that product.

Haha, I missed the addition of the 1, which would make the answer NO, since you're adding one, and the prime number must be at least 3; thus since you have a divisible number by p and add 1, the resultant number is not. So yeah, the answer is no.

[YourLatestVicti]

Why am I not suprised that Ryan has an answer that I have to believe is correct.

[the_narrator]

Oh, and the answer is of course it does, how is that hard to see? (nope, it's not, see addendum).

A prime number is a number that has only itself and one as a divisor, like 7 or 13 or whatever. If you take a number greater than that number (say you take 15, which is greater than 13) and multiply it by every number underneath it until 1, you obviously multiply it by the number p (here 13). Thus the result is divisible by p (13) because the number was a product of p (13) and different numbers. Thus the answer is a whole number equal to m * (m-1) * (m-2) ..... * 1 excluding p itself. Hence p divides the product into some whole number because it was a part of creating that product.

Haha, I missed the addition of the 1, which would make the answer NO, since you're adding one, and the prime number must be at least 3; thus since you have a divisible number by p and add 1, the resultant number is not. So yeah, the answer is no. Do you know why that was so awesome? Because if you didn't read through it...if he had just put "yes" instead of reading it...he'd have gotten it wrong. Perfect. ;) Good job by the way. I can't do math. I've got a calculator & ten friends at Tech school to do that. :)

[MLLMillenium]

OMG I love you.

lol. One last one and I am done for math till next week! YES!

Express the first 15 even numbers greater than 2 as the sum of two prime numbers

Also, your fuckin awesome.

That one really shouldn't be hard....assuming you consider 1 to be prime (which you have to do here, even though it technically is neither)....4: 3 and 1; 6: 5 and 1; 8: 7 and 1 or 5 and 3; 10: 7 and 3; 12: 7 and 5 or 11 and 1; 14: 13 and 1 or 7 and 7; 16: 11 and 5, etc., etc...
UHm.. either A) Im retarded B) I shouldnt be in Math at all.. which is true.. C) Both plus my brain is fired from the math test I had today.. could u PLEASE elaborate a wee bit I will love u forever!!

[Ryan Mills]

That one is an interesting question....Just use the prime numbers... 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, etc. and use them to add up to the number you need.

Just use two of those to add up to each number (4, 6, 8, 10, 12, etc.).

Also what math class is this for...it's so pointless. All this crap they make you do in high school and maybe remedial college math is so tedious for no reason. Calculus owns all that garbage. Advanced statistics too.

[MLLMillenium]

Im in college, not remidial. Anyways uhm college math with stats. It makes no sense. I don't think my teacher doesn't know it's stats cuz we havnt done any!!??

[Ryan Mills]

Im in college, not remidial. Anyways uhm college math with stats. It makes no sense. I don't think my teacher doesn't know it's stats cuz we havnt done any!!??

Yeah, don't worry about it....don't use 1 as a prime, I went through some and never had to use 1, so I'm assuming you'd jusr use all the other primes which I listed above (not all inclusive of course).

Anyways, statistics is pretty basic and easy for intro classes anyways, so I wouldn't expect to see much that's very interesting. You usually don't even get to multiple regression and dealing with heteroskedasticity, serial correlation and other various model problems. Plus, everything will be linear. Anyways, do yourself a favor and take some calculus classes, they're actually useful.

[MLLMillenium]

So wait.. u sure.. cuz thats what I wanted to but it doesnt say the sum of the first 15. SO that would make the answer...? so lost.

[Ryan Mills]

Like for that problem:

4 = 2 + 2
6 = 3 + 3
8 = 5 + 3
10 = 7+ 3
12 = 7 + 5
14 = 11 + 3
16 = 13+ 3
18 = 11 + 7
20 = 17 + 3
22 = 17 + 5
24 = 19 + 5
26 = 23 + 3
28 = 23 + 5
30 = 23 + 7
32 = 29 + 3
And so on...

[MLLMillenium]

Yeah, don't worry about it....don't use 1 as a prime, I went through some and never had to use 1, so I'm assuming you'd jusr use all the other primes which I listed above (not all inclusive of course).

Anyways, statistics is pretty basic and easy for intro classes anyways, so I wouldn't expect to see much that's very interesting. You usually don't even get to multiple regression and dealing with heteroskedasticity, serial correlation and other various model problems. Plus, everything will be linear. Anyways, do yourself a favor and take some calculus classes, they're actually useful.
Oh NO!!!! lol. Last math course of my life. Im studying film as a communications major. Me = No more math. lol.

[Ryan Mills]

So wait.. u sure.. cuz thats what I wanted to but it doesnt say the sum of the first 15. SO that would make the answer...? so lost.

Do you mean the first problem? Think about it. Take 13, which is prime. Then take a number (we call it m here) that's greater, say 15. Now take 15 and multiply it by 14 then that by 13 then that by 12 and so on all the way down till you're multiplying it by 1 (15*14*13*12*11*10*9*8*7*6*5*4*3*2* 1). Now this number IS divisible by 13 (which is p) because it is just a product of a bunch of numbers with 13 being one. So if you divided that by 13, you'd get 15*14*12*11*10....*1 as an answer, which is a whole number. BUT if you add one to that number and then try to divide by 13 it won't be a whole number right? Because if 15*14*13*12*...*1 was divisible by 13, if you add only 1, it would get you a nonwhole number, because you'd need to add 13 to get a whole number (or some multiple of 13). So since the smallest prime number is 2, which is not 1, adding one to the product of all the numbers from 1 till m (a number greater than p) will make it no longer divisible by p because of the addition of 1.

[MLLMillenium]

Do you mean the first problem? Think about it. Take 13, which is prime. Then take a number (we call it m here) that's greater, say 15. Now take 15 and multiply it by 14 then that by 13 then that by 12 and so on all the way down till you're multiplying it by 1 (15*14*13*12*11*10*9*8*7*6*5*4*3*2* 1). Now this number IS divisible by 13 (which is p) because it is just a product of a bunch of numbers with 13 being one. So if you divided that by 13, you'd get 15*14*12*11*10....*1 as an answer, which is a whole number. BUT if you add one to that number and then try to divide by 13 it won't be a whole number right? Because if 15*14*13*12*...*1 was divisible by 13, if you add only 1, it would get you a nonwhole number, because you'd need to add 13 to get a whole number (or some multiple of 13). So since the smallest prime number is 2, which is not 1, adding one to the product of all the numbers from 1 till m (a number greater than p) will make it no longer divisible by p because of the addition of 1.
No no nvm lol. I confused myself. Sorry about that an u did all that work. damn u must be a genius. your awwesome. U deserve like Scene points^2

[Ryan Mills]

No no nvm lol. I confused myself. Sorry about that an u did all that work. damn u must be a genius. your awwesome. U deserve like Scene points^2

Don't worry about it, I'm not really concerned with my scene points. Plus, I enjoy math. It's not that hard, you just have to think logically.

[apoemtothedead]

That question wasn't even that hard.