三個連續數必能被3整除,咁其他連續數又可以比邊d數整除?

According to your questions, we want to prove that:
'n(n+1)(n+2)(n+3) is divisible by 4' and
'n(n+1)(n+2)(n+3)...(n+h-1) is divisible by h, where h is a positive integer' for all integers n.

Let S(n) be the statement 'n(n+1)(n+2)(n+3) is divisible by 4'.
Case 1: n>0
When n=1,
S(1)=(1)(2)(3)(4)=4*6
∴S(1) is true.
Assume that S(k) is true, i.e.
k(k+1)(k+2)(k+3)=4N , where N is an integer
When n=k+1,
S(k+1)
=(k+1)(k+2)(k+3)(k+4)
=k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)
=4N+4(k+1)(k+2)(k+3) (By induction assumption)
=4[N+(k+1)(k+2)(k+3)]
∴S(k+1) is true.
∴By the principle of Mathematical Induction, S(n) is true for all positive integers n.
Case 2: n=0
For n=0,
S(0)=(0)(1)(2)(3)=0=0*4
∴S(n) is true for n=0.
Case 3: n<0
If n is a negative integer, then let n=m-3 , where m≤2.
∴S(n)=S(m-4)
Now we want to prove that 'm(m-1)(m-2)(m-3) is divisible by 4'.
Let S’(m) be the statement 'm(m-1)(m-2)(m-3) is divisible by 4'.
When m=2,
S (2)=(2)(1)(0)(-1)=0*4
∴S’(2) is true.
Assume that S’(k) is true, i.e.
k(k-1)(k-2)(k-3)=4N , where N is an integer
When m=k-1,
S’(k-1)
=(k-1)(k-2)(k-3)(k-4)
=k(k-1)(k-2)(k-3)-4(k-1)(k-2)(k-3)
=4N-4(k-1)(k-2)(k-3) (By induction assumption)
=4[N-(k-1)(k-2)(k-3)]
∴S’(k-1) is true.
∴By the principle of Backward Induction, S’(m) is true for all m≤2, i.e. S(n) is true for negative integers n.
∴Combining all three cases, S(n) is true for all integers n.

P.S. not enough space, see below

2008-07-12 23:39:03 補充：
Let P(n) be the statement 'n(n+1)(n+2)(n+3)...(n+h-1) is divisible by h'.
Case 1: n>0
When n=1,
P(1)=1=1*1
∴P(1) is true.
Assume that P(k) is true, i.e.
k(k+1)(k+2)(k+3)…(k+h-1)=hN , where N is an integer
When n=k+1,
P(k+1)
=(k+1)(k+2)(k+3)(k+4)…(k+h-1)(k+h)

2008-07-12 23:40:06 補充：
=k(k+1)(k+2)(k+3)(k+4)…(k+h-1)+h(k+1)(k+2)(k+3)(k+4)…(k+h-1)
=hN+h(k+1)(k+2)(k+3)(k+4)…(k+h-1)
(By induction assumption)
=h[N+(k+1)(k+2)(k+3)(k+4)…(k+h-1)]
∴P(k+1) is true.
∴By the principle of Mathematical Induction, P(n) is true for all positive integers n.
Case 2: n=0

2008-07-12 23:41:16 補充：
For n=0,
P(0)=(0)(1)(2)(3)…(h-1)=0=0*h
∴P(n) is true for n=0.
Case 3: n<0
If n is a negative integer, then let n=m-h+1 , where m≤h-2
∴P(n)=P(m-h+1)
Now we want to prove that 'm(m-1)(m-2)(m-3)…(m-h+1)(-1)^h is divisible by h'.
Let P'(m) be the statement 'm(m-1)(m-2)(m-3)…(m-h+1)(-1)^h is divisible by h'.

2008-07-12 23:41:49 補充：
When m=h-2,
P’(h-2)=(h-2)(h-3)(h-4)(h-5)…(0)(-1)
∴P’(h-2) is true.
Assume that P’(k) is true, i.e.
k(k-1)(k-2)(k-3)…(k-h+1)(-1)^h=hN , where N is an integer
When m=k-1,
P(k-1)
=(k-1)(k-2)(k-3)(k-4)…(k-h+1)(k-h)(-1)^h
=k(k-1)(k-2)(k-3)(k-4)…(k-h+1) (-1)^h -h(k-1)(k-2)(k-3)(k-4)…(k-h+1)(-1)^h

2008-07-12 23:42:03 補充：
=hN-h(k-1)(k-2)(k-3)(k-4)…(k-h+1)(-1)^h
(By induction assumption)
=h[N-(k-1)(k-2)(k-3)(k+-)…(k-h+1)(-1)^h]
∴P’(k-1) is true.
∴By the principle of Backward Induction, P’(m) is true for all m≤h-2, i.e. P’(n) is true for negative integers n.
Combining all three cases, P(n) is true for all integers n.

2008-07-12 23:42:33 補充：
In fact consecutive numbers is divisible by 8.
Let S(n) be the statement 'n(n+1)(n+2)(n+3) is divisible by 8'.
Case 1: n>0
When n=1,
S(1)=(1)(2)(3)(4)=8*3
∴S(1) is true.
Assume that S(k) is true, i.e.
k(k+1)(k+2)(k+3)=8N , where N is an integer
When n=k+1,
S(k+1)

2008-07-12 23:42:48 補充：
=(k+1)(k+2)(k+3)(k+4)
=k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)
=8N+4(k+1)(k+2)(k+3) (By induction assumption)
=8N+24M (Since (k+1)(k+2)(k+3)=6M for some integer M)
=8(N+3M)

2008-07-12 23:42:59 補充：
∴S(k+1) is true.
∴By the principle of Mathematical Induction, S(n) is true for all positive integers n.
For the case n≤0 they are similar like above.

2008-07-12 23:43:30 補充：
The logic is as follow:
For any three consecutive numbers, e.g. {1, 2 and 3}, {100, 101 and 102}, …
There must exist a factor of 2 and a factor of 3, i.e. the product of them must has a factor of 6 (2*3=6).

2008-07-12 23:43:41 補充：
Similarly, for any four consecutive numbers, there must exist a factor of 2, 3 and 4, i.e. the product of them must contains a factor of 24 (2*3*4=24).
Therefore, the product of n consecutive numbers is not only divisible by n but divisible by (1)(2)(3)…(n), i.e.
∏(r=1 to n)r.

考慮 k 個連續數乘埋，張呢k個連續數都除以k，佢地既餘數冇可能一樣，這是因為兩個除k餘數一樣既數差一定能整除k。

例子, 2, 3, 4, 5 除4既餘數分別就係 2, 3, 0, 1
例子, 7, 8, 9, 10, 11 除5既餘數分別就係 2, 3, 4, 0, 1

除k 既餘數只可能係由0, 1, ... , k-1，而家有k個唔同既餘數，而呢k個餘數裏面又一定只可以係由0到k-1，由此可知呢k個餘數裏面一定有0。

所以k個連續數裏面，必然有一個k既倍數，所以它們的積一定可以整除k。

唔係咩都要M.I. 既。

Think about the divisibility of integers

先用簡單等式想一下 (不是什麼都用數學歸納法的嘛)

好似4個連續數(n, n+1, n+2, n+3)

一加起來 就知和 = 4n+6, 4n是4的倍數 但是6不是4的倍數 所以不能被4整除

而 5個連續數 (n,n+1,n+2,n+3,n+4) = 5n+10 = 5(n+2) 可以被5整除

......


那麼請你想想

如果n的總數是奇數的話 是不所有n都成立?
又如果是n總數是偶數的話 是不是所有n都不成立?

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知道答案未必就是全部知識 思考的過程也很重要
這題的分數於我不重要

2008-07-19 22:06:29 補充：
我以為你說的是sum of the numbers 原來是product 那麼當我沒答過 不好意思