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# 兩到高一數學證明題

1.N為小於 2003之 正整數 (n平方-13)除以(n+5) 為最簡分數

N值有幾個?

2.有 A , B 二正整數

請證明: AB+1 與 A 互質

to the second solver:

why is the general formula

(N+5)-[(N+5)/2]-[(N+5)/3]+[(N+5)/6]-2

i don't understand , please explain!

### 2 Answers

- myisland8132Lv 71 decade agoFavorite Answer
1 Consider (n^2-13)/(n+5) if it is in simplest form. Then

(n+5,n^2-13)=1=>(n+5,n^2-13-n(n+5)=1

=>(n+5,5n+13)=1=>(n+5,5n+13-5(n+5))=1=>(n+5,12)=1

Since 12=2*2*3, we conclude that n+5 is coprime with 2 and 3

From 0<n<2003, the number of n+5 which is coprime with 2 and 3 is

2007-[2007/2]-[2007/3]+[2007/6]-2 (since we omit 1,5)

=2007-1003-669+334-2

=667

There are 667 different values of n

The general formula is

(N+5)-[(N+5)/2]-[(N+5)/3]+[(N+5)/6]-2

2 if this is not true, then AB+1=MA=>AB-MA=-1

=>A(M-B)=1

That means the product of two integers is equal to 1, which is a contradiction !

2009-08-16 12:34:04 補充：

For Q.2, we assume A>1 !

- 高萌度數學娘Lv 61 decade ago
1.(n^-13,n+5)=1

(n^-13)/(n+5)=n+(-5n-13)/(n+5)

=n-5+[12/(n+5)]

所以若此為最簡分數，則n+5不為2，3之倍數

故所求=1335個

2.設(ab+1，a)=d

則d l ab+1且d l a

d l ab+1-ab

d l 1

d=1

故ab+1與a互質

算錯請指正

Source(s): 我自己+數學娘的加持