prove or disprove: the product of any four consecutive integers is divisible by 24. From what I figure, this is true, but I'm not sure how to prove it. Haven't found a counterexample.
let a be the first integer: a*(a+1)*(a+2)*(a+3)
what does "is divisible" meant to result in?
or can we do this by induction.. amistre thoughts?
an integer
n/24 is valid for all n if we are just concerned with real values
to disprove you anly need one example thats false
as in n/24 is an integer. and i haven't found a coutnerexample yet to disprove it
-1-2 1 2 = 4 4/24 is not an integer
Oh, sorry only the positive integers...
those aint consecutive lol
4 consecutive integers i thought starting from 1
Induction is for kidds :P
any number less than 24 is going to go off 1*2*3*4/6*4 = 1 2*3*4*5/4*3*2 = 5 3*4*5*6/6*4 = 15 induction, yep
ffm it may be for kids, but it helps
For divisibility by 24 you need to only show the divisibility by 3 and 4 (Excercise: why?) Now, any 4 consecutive integer must have at-least 2 even numbers and at-least one multiple of 3 hence ... QED
(2n-1)(2n)(2n+1)(4n) 8n^2(4n^2 - 1) 32n^4 - 8n^2 ------------------ = -------------- = ------------ 24 24 24
4n^4 - n^2 ---------- looks less hairy 3
For an inductive proof see here http://www.algebra.com/algebra/homework/word/numbers/Numbers_Word_Problems.faq.question.478538.html
factor out an n^2 to get: \[n^2(\frac{4n^2-1}{3})\] since n^2 is always an integer we just got to focus on the other :)
since 4n^2 - 1 = (2n-1)(2n+1) all we have to do is find a way to prove that the product of 2 consecutive odd intergers is divisible by 3 n=1 ; 1*3 = 3 n=2 ; 3*5 = 15 n=3 ; 5*7 = 35 , but since n^2 = 9 is /3 then we are safe n=4 ; 7*9 = 63 n=5 ; 9*11 = 99 n=6 ; 11*13 = 143 ; but again, n^2 = 36 so it turns out fine again. whenever n is a multiple of 3 we get an automatic ok since n^2 is divisible by 3 but i got no idea how to write up a proff for it :/
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