Olympic Mathematics
本帖最後由 -終場ソ使者- 於 28-11-2009 00:50 編輯一齊討論下:
1.Prove that for any integer n,n6+2n5-n2-2n can be divisible by 120.
2.Given that for any a belong to real number, let the sum of a1,a2,...,an can be divisible by 30,prove that a15,a25,...,an5 can also be divisible by 30.
(1.證明對任意整數n,n6+2n5-n2-2n 能被120整除.
2.設a1,a2,...,an 是自然數,它們的和能被30整除,證明a15,a25,...,an5 的和也能被30整除.) 本帖最後由 wong87 於 29-11-2009 11:41 編輯
m2?
整除性?
數學歸納法?
1.證明對任意整數n,n6+2n5-n2-2n 能被120整除.
2.設a1,a2,...,an 是自然數,它們的和能被30整除,證明a15,a25,...,an5 的和也能被30整除.)
1.
設p(n)為命題6+2n5-n2-2n可被120整除]
當n=1, 16+2(1)5-(1)2-2(1)=1+2-1-2=0
p(1)成立
設p(k)成立,k6+2k5-k2-2k ,k6=-2k5+k22+k
當n=k+1
(k+1)6+2(k+1)5-(k+1)2-2(k+1)
跟住就要靠自己啦:) 本帖最後由 bunbunbunbun 於 21-12-2009 00:02 編輯
一齊討論下:
1.Prove that for any integer n,n6+2n5-n2-2n can be divisible by 120.
2.Given that for any a belong to real number, let the sum of a1,a2,...,an can be divisible by 30,prove that a15,a25,... ...
-終場ソ使者- 發表於 27-11-2009 16:47 http://www.nakuz.com/bbs/images/common/back.gif
1. n^6+2n^5-n^2-2n=n(n^5+2n^4-n-2)=n(n+2)(n^4-1)=n(n+2)(n^2-1)(n^2+1)=n(n+2)(n-1)(n+1)(n^2+1)=(n-1)(n)(n+1)(n+2)(n^2+1)
Since n-1, n, n+1 and n+2 are four consecutive integers, one of them must be divisible by 2,one of them must be divisible by 3 andone of them must be divisible by 4.
Then if the one of the four integers is divisible by 5, (n-1)(n)(n+1)(n+2)(n^2+1) is divisible by 5
If all of them are not divisible by 5, then n = 2 mod 5
n^2 = 4 mod 5
n^2+1 = 0mod5, i.e. n^2+1 is divisible by 5
Thus the statement is true for all integers
NB, this Q cannot be done by simple induction as you have to prove that this is true for all integers, not just positive. (or you can do two inductions i suppose....)
2. real numbers or natural numbers-.-? Plus I haven't learn about cyclic summation, not sure if I can do that
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