理科人、a.maths人請進！！超緊急!! [復制鏈接]

呢幾條點計??

is it not a.maths =.=?

i haven't seen these kind of sentences before..

等版主解答.......

a maths pure maths 都無e d數架喎.....

我try my best答啦

1. find all n s.t. n^2+13 is divisible by 14

n 一定要係odd先啦

n=1 ok

now let n^2+13 = 14k, k is an integer

n^2-1 + 14 = 14k


(n^2-1)/14 + 14/14 = k

(n^2-1)/14 = k-1

(n+1) (n-1)/14 = k-1

So (n+1) (n-1)/14 needa be an integer

So solve all n's s.t. n+1=14m or n-1=14p, where p, m are integers
i.e. Get n=13,27,41, ... or n=1,15,29,43,... (Q.E.D)

q2, 3, 4, 5 may take a ref here:
http://residence.educities.edu.tw/kuen/project/divide.htm

http://tw.knowledge.yahoo.com/question/?qid=1105062007957

4. 4x=7(mod 9), 100<x<200,
4x=7+9m, m is an integer

400<4x=7+9m<800
43.77<m<88.11
(So The range of m taken gotta be between 44 and 88)

4x=7+9m=8(1+m)+(m-1)

x=2(1+m) + (m-1)/4
So needa take m s.t. m-1 is divisible by 4 and in the range from 44 to 88

i.e. m= 45, 49, 53, ..., 85
Sub these into 4x=7+9m, get x= 103, 112, 121, ..., 193


For Qs 2,3,5 I really donno la.....but just take a look at the ref links above

首先非常感謝版主的努力

真的非常非常感謝

但是...我仍有不明白的地方啊......

Get n=13,27,41, ... or n=1,15,29,43,... (Q.E.D)
即係咩??

點解我唔識既..ED應該幾時教嫁= . = "

下面是引用迪克於2005-10-30 21:46發表的:
首先非常感謝版主的努力
真的非常非常感謝
但是...我仍有不明白的地方啊......
Get n=13,27,41, ... or n=1,15,29,43,... (Q.E.D)
即係咩??

Get n=13,27,41, ... or n=1,15,29,43,...

姐係 n=13, 27, 41, 等等(每個term加14) 同埋 1,15,29,43, 等等(每個term加14)

下面是引用啊謝於2005-10-30 22:12發表的:
點解我唔識既..ED應該幾時教嫁= . = "

都話a maths, pure maths 都無囉

唔該你呀....

我睇左d web..都唔太明點做2,3,5...
有冇高手可以幫幫手...

呢d係 number theory來的...

原來仲有一題...........
解6x=8(mod22)

下面是引用迪克於2005-10-31 01:33發表的:
原來仲有一題...........
解6x=8(mod22)

同Q4原理一樣, 你先試下自己做

下面是引用kylau於2005-10-31 01:46發表的:
同Q4原理一樣, 你先試下自己做

試過....
卓到x=(4+11m)/3

好似唔太o岩...

我咁做：
6x=8mod22
6x=8+22m
6x=2(4+11m)
3x=4+11m
x=(4+11m)/3.........

oh for the q4, i've assumed x must be an integer. But, if that is not required, we then have missed some answers

in fact, if the condition that "x must be an integer" is not required, then m can be any integer from 44 to 88 as we've calculated 43.77<m<88.11

i.e. m=44, 45, ..., 88
sub into 4x=7+9m, get x=100.75, 103, 105.25, ..., 199.75 (每項加2.25)

--------------------------------------
6x=8mod22
6x=8+22m
6x=2(4+11m)
3x=4+11m=6+9m+2m-2

x=(1+3m) + 2(m-1)/3

後面(m-1) 要divisible by 3
take m=3k+1, k is an integer

你e條個 x 無bounded, 所以係有無限個solution,
sub m=3k+1 into 3x=4+11m,

get x=11k+5 where k is an integer

thanks~
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