高中數學：整數論的問題

Question

在自然數中，使2^(3n+3)-7n+41恆為m的倍數，則最大的m是什麼？

問題是可以解，但大家的解法會不同，看看大家的解法也不錯呀！
再等等是否還有別的寫法。

都沒有人再有解答了，那我就讓他們交付投票，由於三號的方法和一號相似，所以我就採一號(先答)和二號兩種方法來，讓網友們決定什那一種較好。

Answer 1

Answer: 最大的 m 是 49.

Let A(n) = 2^(3n + 3) - 7n + 41.Then

A(1) = 7^2*2 is a multiple of 49.

A(2) = 7^2*11 is a multiple of 49.

A(3) = 7^3*2^2*3 is a multiple of 49.

Also, note that the g.c.d. of A(1), A(2) and A(3) is 49.

So let's claim that the maximum possible m is 7^2 = 49.

We made an induction on n to complete the proof.

For n=1: A(1) = 98 = 49*2, it is clear holds.

Suppose A(n) = 2^(3n + 3) - 7n + 41= 49 r, r be an integer.

Then 2^(3n+3) = 49r + 7n - 41.

By using the induction hypothesis we have

A(n+1) = 2^(3n+6) - 7n - 7 + 41 = 8*2^(3n + 3) - 7n - 7 + 41

=8* (49r + 7n - 41) - 7n - 7 + 41 = 49 (8r + n - 6) is a multiple of 49.

So A(n) is a multiple of 49, so does A(n+1).

Done!

2007-12-27 10:27:37 補充：
To tsl:


By looking your problem-sloving credit, I think you are able to solve such

a easy question!

So, I am confused why you raised such a problem.

Answer 2

Ans: 49

(1)n=1, n=2時均為49倍數
(2)原式=8n+1-8-7n+49
=(1+7)n+1 -8-7n+49
=1+(n+1)7+C(n,2)72+... - 8 - 7n+ 49
=49C(n,2)+...+49
=49倍數


2007-12-27 16:22:52 補充：
To: wei
提出問題可能有多面向的想法,大部份純粹是求知,困擾; 也可能想看看或提供討論的話題; 也可能覺得這題目很有意思,給大家欣賞;...

這個題目算是特別,簡簡單單的數學式,居然是49的倍數,能看到7的倍數就值得嘉獎了,能證明更好.
本題應該是高一數學歸納法很好練習,不好意思,我無端以二項式展開解說,高中生若看不懂,請多包涵啦!