Ugh I am confused over tiny technical details..(Original post by Matureb)
No. In this case 30 is divisible by 2 and by 6. But not by 18. For 18 you would need to show it was divisible by 2 and 9.
So to prove something is divisible by 30, it would be 15 and 2 ?
for 18  2 and 9
for x  any two factors which multiply to give x ?
Or is it that one of the factors must always be 2 ?
eg to prove something is divisible by 60 can I use 10 and 6 or must I use 2 and 30, or will 4 and 15 also work ?

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 21
 16042012 15:40

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 22
 16042012 15:41
(Original post by member910132)
Oh, can we say that at least one of those numbers must be even and so the whole thing can be divided by 2 and so the whole number is even.
But can someone clarify this issue of using factors to prove something is divisible by a number ?
If I wanted to prove something is divisible by x, then obviously proving it is divisible by Ax would suffice right ? (where A is a constant integer).
But if I am to prove that something is divisible by 12 for example then must I say that it is divisible by 2 of it's factors, eg show that it is divisible by both 3 and 4 or 2 and 6 ?
Thnx tons 
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 23
 16042012 15:55
(Original post by miml)
In general if you want to prove x is divisible by y then you have to show x is divisible by each of the prime factors of y. 
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 16042012 16:00
60 = 2*2*3*5. To test if a nunber is divisible by 60 it must be divisible by (2*2), by 3 and by 5. I.e it must be divisible by 3, 4 and 5. Note, these are relatively prime.
You are right that 4 and 15 will do since if it is divisible by 15, then it is automatically divisible by 3 and 5.
10 and 6 would not do. 90 is divisible by 10 and 6 , but not by 60. 
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 25
 16042012 16:04
Surely it would be easier to say
1) when n=k
assume n^3n is divisible by 6
2) when n=k+1
n^3+3n^2+2n
1)x4 + 2)x2 gives 4n^34n + 2n^3+6n^2+4n = 6n^3+6n^2 = 6(n^3+n^2)
......................(divisible by 6).............................. .................(divisible by 6)
As the two bits I've shown are divisible by 6, the 6n^2+4n must be divisible by 6 and so is the equation in 2). 
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 26
 16042012 19:05
(Original post by Matureb)
60 = 2*2*3*5. To test if a nunber is divisible by 60 it must be divisible by (2*2), by 3 and by 5. I.e it must be divisible by 3, 4 and 5. Note, these are relatively prime.
You are right that 4 and 15 will do since if it is divisible by 15, then it is automatically divisible by 3 and 5.
10 and 6 would not do. 90 is divisible by 10 and 6 , but not by 60.
(Original post by Quip)
Surely it would be easier to say
1) when n=k
assume n^3n is divisible by 6
2) when n=k+1
n^3+3n^2+2n
1)x4 + 2)x2 gives 4n^34n + 2n^3+6n^2+4n = 6n^3+6n^2 = 6(n^3+n^2)
......................(divisible by 6).............................. .................(divisible by 6)
As the two bits I've shown are divisible by 6, the 6n^2+4n must be divisible by 6 and so is the equation in 2). 
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 27
 16042012 19:25
(Original post by Quip)
Surely it would be easier to say
1) when n=k
assume n^3n is divisible by 6
2) when n=k+1
n^3+3n^2+2n
1)x4 + 2)x2 gives 4n^34n + 2n^3+6n^2+4n = 6n^3+6n^2 = 6(n^3+n^2)
......................(divisible by 6).............................. .................(divisible by 6)
As the two bits I've shown are divisible by 6, the 6n^2+4n must be divisible by 6 and so is the equation in 2).
Surely you want 2n^3 + 6n^2 + 4n to be divisible by 12, to show that our original expression is divisible by 6. 
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 28
 16042012 20:14
(Original post by miml)
Might be missing something, but your argument only shows that the "equation 2" is divisible by 2?
Surely you want 2n^3 + 6n^2 + 4n to be divisible by 12, to show that our original expression is divisible by 6.
However, I've just realised that as I've already multiplied it by 2, this only shows that n^3 + 3n^2 + 2n is divisible by 3 and you're right I need to show 2n^3 + 6n^2 + 4n is divisible by 12. Sorry, my mistake (I've been on holiday too long and forgotten how to do maths) 
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 29
 16042012 20:55
there are different cases to consider:
firstly, n^3n can be expressed as: n(n1)(n+1)
so, first, if n is divisible by 3, then n1 is divisible by2, so whole thing divisible by 6  nothing further needed there.
second thing you do is examine what happens when n is both odd and even: odd n=2k+1 for some k in Z  you find out, by factoring, that both expressions are divisible by 2,
3rd, you examine n when it`s NOT divisible by 3  by using in the expression, (then factoring it) both n=3k+1, and n=3k1 (this ensures thatyou encompass every number NOT divisible by 3)  you find that the expression n^31 when used with these values, simplifies to something with a factor of 3,
so you will have proved simultaneously that the expression has factors of both 2 and 3...Last edited by Hasufel; 16042012 at 20:57. 
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 30
 16042012 21:15
(Original post by Hasufel)
there are different cases to consider:
firstly, n^3n can be expressed as: n(n1)(n+1)
so, first, if n is divisible by 3, then n1 is divisible by2, so whole thing divisible by 6  nothing further needed there.
second thing you do is examine what happens when n is both odd and even: odd n=2k+1 for some k in Z  you find out, by factoring, that both expressions are divisible by 2,
3rd, you examine n when it`s NOT divisible by 3  by using in the expression, (then factoring it) both n=3k+1, and n=3k1 (this ensures thatyou encompass every number NOT divisible by 3)  you find that the expression n^31 when used with these values, simplifies to something with a factor of 3,
so you will have proved simultaneously that the expression has factors of both 2 and 3...
To prove by induction that a number is divisible by some number, the difference between k and k+1 must be used. Otherwise its just a direct proof.
(Original post by BJack)
.Last edited by Intriguing Alias; 16042012 at 21:18. 
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 31
 16042012 21:31
How I would have done it, if it had to be done by induction:
Assume true for n=k then is divisible by 6.
For n=k+1
The first bracket is divisible by 6 by our assumption so it remains to show the second bracket is.
Which is clearly divisible by 3, and as it contains a product of two consecutive numbers i.e. one must be even, it is also divisible by 2. Hence divisible by 6 and we have shown that for n=k+1 the expression is divisible by 6.
If it didn't have to be done by induction
Spoiler:Shown^3  n = (n1)n(n+1) which is the product of 3 consecutive integers, hence is divisible by 3 and 2 
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 32
 17042012 08:48
(Original post by hassi94)
Also a problem with your idea of just showing either number is a multiple of 6 is that it isn't induction. You can do all the induction steps to disguise it as induction, but unless you actually use the assumption that n³  n is a multiple of 6 then its just a direct proof.
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Updated: April 17, 2012
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