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#1 2009-03-14 01:48:55

JaneFairfax
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#2 2009-03-14 05:49:38

JaneFairfax
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#3 2009-03-14 06:48:46

JaneFairfax
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#4 2009-03-14 11:27:58

JaneFairfax
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#5 2009-03-15 05:03:36

JaneFairfax
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#6 2009-03-16 20:10:55

JaneFairfax
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[align=center]

[/align]

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#7 2009-03-16 20:15:48

JaneFairfax
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#8 2009-03-18 06:45:10

JaneFairfax
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#9 2009-03-21 12:09:22

Muggleton
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#10 2009-03-21 12:43:58

JaneFairfax
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Yup!

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#11 2009-03-21 22:14:29

JaneFairfax
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Re: Janes Exercises III

Last edited by JaneFairfax (2009-03-22 03:33:42)

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#12 2009-03-22 03:10:46

mathsyperson
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Re: Janes Exercises III

d=e=1 disproves both parts of #9.

(But not anymore.)

Why did the vector cross the road?
It wanted to be normal.

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#13 2009-03-22 03:41:17

JaneFairfax
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Re: Janes Exercises III

I misunderstood the term square free. I thought it was synonymous with non perfect square (and therefore excluded 1).

I have rephrased the question. It shouldnt be ambiguous now.

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#14 2009-03-29 00:40:29

JaneFairfax
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#15 2009-03-29 23:11:41

JaneFairfax
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#16 2009-03-30 00:03:19

Kurre
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#17 2009-03-30 00:34:41

smiyc86
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Re: Janes Exercises III

why has no one tried #1 ???

newayz ...

we can rewrite the equn as ... (n-2)(n-1)(n²-2)(n+1)(n+2)

we can clearly see that this expression will be div for n= 7k+1,7k+2,7k+5,7k+6 ... as in each of these casesone of n-1,n-2,n+1,n+2 ... contributes a factor of 7

for 7k+3 and 7k+4 we can find, by substitution, that n²-2 donates a factor of 7.

hence the expression is div by 7 ...

now, if n-1 and n+1 are even and as they are consecutive nos. we can write them as ... 2a,2a+2
now, 2a*(2a+2) = 4a*(a+1)
now a(a+1) will be div by 2! hence div of 8 is also true when n-1 is even

now if n-2 is even ... so is n²-2 and n+2 ....
3 even nos hence three multiples of 8 ... hence div by 8 ....

thus the exp is always div by 56 ...

I love Maths and Music ... dunno which more

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#18 2009-03-30 10:23:50

JaneFairfax
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@Kurre
Well done!

@smiyc86

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#19 2009-03-30 11:25:33

JaneFairfax
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#20 2009-03-30 17:30:10

smiyc86
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Re: Janes Exercises III

@jane ... how to use the hidden text ??? some html codes reqd if so give the name of the starting tag ... btw i was going o use 7k-1 and 7k-2 instead of 7k+6 and 7k+5 .... but somehow didn't

I love Maths and Music ... dunno which more

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#21 2009-03-30 17:34:34

smiyc86
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Re: Janes Exercises III

I love Maths and Music ... dunno which more

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#22 2009-03-30 18:50:56

smiyc86
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Re: Janes Exercises III

Last edited by smiyc86 (2009-03-30 19:52:55)

I love Maths and Music ... dunno which more

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#23 2009-03-30 19:13:08

smiyc86
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Re: Janes Exercises III

Last edited by smiyc86 (2009-03-31 18:32:37)

I love Maths and Music ... dunno which more

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#24 2009-03-30 22:54:27

JaneFairfax
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Re: Janes Exercises III

smiyc86:

#6. Brilliant!

#9. Please expand on the proof.

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#25 2009-03-31 18:34:43

smiyc86
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Re: Janes Exercises III

@9 ... done

I love Maths and Music ... dunno which more

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