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Prove 16x(4x+5)(x+1) is a multiple of 48 where x is greater than or equal to 1.

Can someone prove this?

16x(4x+5)(x+1) is a multiple of 48 where x is greater than or equal to 1.

This should be obvious but its not - ive checked it up to x = 10 and it works.

Thanks in advance!
Original post by bnosnehpets
Can someone prove this?

16x(4x+5)(x+1) is a multiple of 48 where x is greater than or equal to 1.

This should be obvious but its not - ive checked it up to x = 10 and it works.

Thanks in advance!


x has to be one of

3n
3n+1
3n+2

since all numbers are
Reply 2
Original post by TenOfThem
x has to be one of

3n
3n+1
3n+2

since all numbers are


All numbers are not one of 3n, 3n+1 or 3n+2 (unless n can be 1/3, 2/3, etc.). In that case surely you could say that for any number of ns. For example x is one of 100n, 100n+1, 100n+2, 100n+3, ... and prove that it is a multiple of 1600?

This doesn't make sense to me, can you explain more?
Original post by bnosnehpets
All numbers are not one of 3n, 3n+1 or 3n+2 (unless n can be 1/3, 2/3, etc.). In that case surely you could say that for any number of ns. For example x is one of 100n, 100n+1, 100n+2, 100n+3, ... and prove that it is a multiple of 1600?

This doesn't make sense to me, can you explain more?


All integers are of the form 3n, 3n+1, 3n+2

I have assumed that x is an integer - otherwise your question does not actually make sense

Not sure what you mean with the 1600 but as that would not work
(edited 9 years ago)
Reply 4
Original post by TenOfThem
I have assumed that x is an integer - otherwise your question does not actually make sense

Not sure what you mean with the 1600 but as that would not work


Sorry, I should have specified. x is an interger. Yes, I now understand what you mean by all intergers are of form 3n, 3n+1, 3n+2. However are all intergers not also of the form 4n, 4n+1,4n+2,4n+3 and 5n,5n+1 etc...

I don't understand how this effects the question though. Could you please explain in a bit more detail how what you said effects the proof?

thanks
(edited 9 years ago)
Original post by bnosnehpets
Sorry, I should have specified. x is an interger.

I still don't understand though. could you please explain in a bit more detail how what you said effects the proof?

thanks


Have you tried putting the possibilities into the original expression

All should become clear
Reply 6
Original post by bnosnehpets
Sorry, I should have specified. x is an interger.

I still don't understand though. could you please explain in a bit more detail how what you said effects the proof?

thanks


If x is an integer and you divide it by 3, there are 3 possible results:

remainder 0 i.e. x is divisible by 3 i.e. x = 3n for some integer n
remainder 1 i.e. x = 3n + 1 for some integer n
remainder 2 i.e. x = 3n + 2 for some integer n
Sounds like a proof by induction, given the condition on x...
Original post by alexmufc1995
Sounds like a proof by induction, given the condition on x...


My method is much simpler
Reply 9
Original post by alexmufc1995
Sounds like a proof by induction, given the condition on x...

hmm ok, that seems strange considering this is not in that section of my course
Original post by TenOfThem
My method is much simpler


Original post by bnosnehpets
hmm ok, that seems strange considering this is not in that section of my course


If you don't know it, try TenOfThem's method. It's more intuitive anyway.
Original post by TenOfThem
Have you tried putting the possibilities into the original expression

All should become clear

Yes, I have tried up to x=10 and all work.
But I still don't understand how you can prove this. Please can someone talk me through the steps they would take?

Edit: Are you telling me to sub 3n, 3n+1 and 3n+2 into the equation and see if they come out with multiples of 48?
(edited 9 years ago)
Original post by bnosnehpets
Yes, I have tried up to x=10 and all work.
But I still don't understand how you can prove this. Please can someone talk me through the steps they would take?


SO you have not put the possibilities in


The steps I would take

Realise that 3 x 16 is 48
Realise that all integers are of one of these forms
3n
3n+1
3n+2
Put these options into the original statement
Look for a factor of 3 in a bracket
Show the 48 is a factor
Original post by TenOfThem
SO you have not put the possibilities in


The steps I would take

Realise that 3 x 16 is 48
Realise that all integers are of one of these forms
3n
3n+1
3n+2
Put these options into the original statement
Look for a factor of 3 in a bracket
Show the 48 is a factor


Ok thank you! I now understand and all forms work.
Original post by bnosnehpets
Ok thank you! I now understand and all forms work.


:biggrin:
Original post by TenOfThem
:biggrin:

In a rigorous proof, would I have to specify that n>=1 ?
Original post by bnosnehpets
In a rigorous proof, would I have to specify that n>=1 ?


hmmmm interesting question

I would say that is implied by the fact that x is greater than or equal to 1
Original post by TenOfThem
hmmmm interesting question

I would say that is implied by the fact that x is greater than or equal to 1

Ok thanks, sorry for the silly comments earlier then, I was overlooking the fact that n could also = 0 (when x = 3n+1 and when x = 3n+2, but not when x = 3n)

Thanks for all your help :smile:
Original post by bnosnehpets
Ok thanks, sorry for the silly comments earlier then, I was overlooking the fact that n could also = 0 (when x = 3n+1 and when x = 3n+2, but not when x = 3n)

Thanks for all your help :smile:


No Problem

Glad that you have understood:biggrin:

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