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    At Southampton today I was given a questions sheet. I've found 2 of the proof by induction questions particularly difficult:

    1. 10^n + 48(4^n)5 is divisible by 9 <<<I can;t get a place to start, have they made a mistake on the worksheet?

    2. x^5 - 5x^3 + 4x is divisible by 120 factorised this so: x(x+1)(x+2)(x-1)(x-2) then let x+3 to show its true for 1 case. not really sure where to go from there?


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    (Original post by ssmoose)
    At Southampton today I was given a questions sheet. I've found 2 of the proof by induction questions particularly difficult:

    1. 10^n + 48(4^n)5 is divisible by 9 <<<I can;t get a place to start, have they made a mistake on the worksheet?
    Is the 5 supposed to be there? If so it doesnt work for n=1.

    2. x^5 - 5x^3 + 4x is divisible by 120 factorised this so: x(x+1)(x+2)(x-1)(x-2) then let x+3 to show its true for 1 case. not really sure where to go from there?
    (x-2)(x-1)x(x+1)(x+2) is the product of 5 consecutive positive integers (supposing x>2. Hence at least one of these factors is divisible by 2, one by 3, one by 4 and one by 5. Call these factors 2a, 3b, 4c, 5d and let the unused factor just have value e (a,b,c,d,e are all arbitrary integers). Then you have x^5-5x^3+4x=e(2a)(3b)(4c)(5d)=120(ab cde). Thus we see it is divisible by 120.
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    (Original post by ssmoose)
    At Southampton today I was given a questions sheet. I've found 2 of the proof by induction questions particularly difficult:

    1. 10^n + 48(4^n)5 is divisible by 9 <<<I can;t get a place to start, have they made a mistake on the worksheet?

    2. x^5 - 5x^3 + 4x is divisible by 120 factorised this so: x(x+1)(x+2)(x-1)(x-2) then let x+3 to show its true for 1 case. not really sure where to go from there?


    Cheers for helping
    hmm. strange as n(1) doesn't work. did they not give any conditions?
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    (x-2)(x-1)x(x+1)(x+2) is the product of 5 consecutive positive integers (supposing x>2. Hence at least one of these factors is divisible by 2, one by 3, one by 4 and one by 5. Call these factors 2a, 3b, 4c, 5d and let the unused factor just have value e (a,b,c,d,e are all arbitrary integers). Then you have x^5-5x^3+4x=e(2a)(3b)(4c)(5d)=120(ab cde). Thus we see it is divisible by 120.
    thankyou! The worksheet says the 5 is there, but it doesn't work even if the 5 is absent does it?

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    (Original post by ssmoose)
    thankyou! The worksheet says the 5 is there, but it doesn't work even if the 5 is absent does it?

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    No conditions were given
    could be a typo, +5 maybe?
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    could be a typo, +5 maybe?
    So far it looks more achievable
 
 
 
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