prove that n^3-9n^2+20n is divisible by 6 where n is greater or equal to 1
asked Apr 24, 2013 in Algebra 2 Answers by anonymous

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Factor the given expression.   We have: n³-9n²+20n=n(n-4)(n-5) ··· Ex.1

If Ex.1 is divisible by 6, Ex.1 is also divisible by two prime factors of 6, 2 and 3. (6=2x3)

A. If n is odd, (n-5) is even.   If n is even, (n-4) is also even.   Therefore, n(n-4)(n-5) is always a multiple of 2.

B. If Exp.1 is divisible by 3, the remainder is 0,1,or 2.

If n ≡ 0 (mod 3), n-4 ≡ -4 ≡ -1 (mod 3), and n-5 ≡ -5 ≡ -2 (mod 3), that is: n is a multiple of 3.

If n ≡ 1 (mod 3), n-4 ≡ -3 ≡ 0 (mod 3), and n-5 ≡ -4 ≡ -1 (mod 3), that is: (n-4) is a multiple of 3.

If n ≡ 2 (mod 3), n-4 ≡ -2 (mod 3), and n-5 ≡ -3 ≡ 0 (mod 3), that is: (n-5) is a multiple of 3.

Therefore, n(n-4)(n-5) is always a multiple of 3.

CK: If n=1, n(n-4)(n-5)=1(-3)(-4)=12, 12(n=2), 6(n=3), 0(n=4), 0(n=5), 12, 42, 96, 180 … CKD. 

Therefore, n³-9n²+20n is divisible by 6 for all integers greater than or equal to 1.

 

answered Jun 4, 2014 by tokiokid70
edited Jun 4, 2014
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