【数学讨论101-200】4.10起酥雪讨论稿（至200题）

kittybite

157 条件2
Proof:
suppose every prime factor of n exists in set P. (we are trying to prove this is not true by reaching a contradiction.)
let P = {p1, p2,....pm}.
now we know n=p1*p2*....*pm+1, this means n is greater than 1, and the greatest common divisor between n and p1*p2*...*pm is 1.
let's write n =( pa^x1)*(pb^x2)*...*(pk^xk). where pa, pb,... px are prime factors of n， x1,x2,...xk are exponents of prime factors of n.

note that we assumed that all prime factors of n are in set P, so pa is in set P, pb is in set P,... pk is in set P.

now let's look at the greatest common divisor between n=( pa^x1)*(pb^x2)*...*(pk^xk), and p1*p2*...*pm. Since pa is in set P, pb is in set P, ... px is in set P, the greatest common divisor between n and p1*p2*...*pm is pa*pb*...*px. and since n is >1, pa*pb*...*px is also greater than 1, therefore we reached a contradiction.

Thus there exists a prime factor of n such that this prime factor is not in set P.
感觉有点奥数做证明了哈哈。

hummergre

关于157 第二个条件，我的理解是：如果p set is p1, p2, p3 ...pk, then n = p1p2p3..pk+1, 如果n的所有质因子都在p set 里面，那么 n 除以p set 里面属于n的任意一个质因子 （say pj, j 属于1 到 k)应该是整除, 但是 n/pj 并不是整除（j 从1 取到k 都无法满足），因为会有一项 1/pj, pj 是质因子，绝对不会是1，所以n/pj无法整除，也就是说n的质因子都不在p set 里。

西域小狼

hummergre 发表于 2016-4-16 09:14
关于157 第二个条件，我的理解是：如果p set is p1, p2, p3 ...pk, then n = p1p2p3..pk+1, 如果n的所有质 ...

NN 好棒！！

shirleyyunun

第112题求分析

jewellai

亲，我更新了128题的，请找一下我发的帖子

AGIRLNAMEDXIN

#124
12点半开始响了第一次，十五分钟响一次，那不就是响了2次而已嘛？为什么是3次呢？求解！

whoseapple

170题错了吧 y轴右边3^x在5^y的下面，画图可以看出来，当xy都大于0时，x是必须大于y的，所以综合条件1，2可以确定xy的关系 不能选E

Branyy

考古：第113题是去年六月库的138题~

whoseapple

还有176题是不是写错了？第二个条件化简可得42/k余12是怎么做到的？不是42/k余6么? 那么条件2满足K的值就有9，12，18，36 最后答案倒确实是a

whoseapple

还有178题我并不是很懂，如果原题说an是奇数 那整个an数列都是奇数了 中位数肯定也是奇数啊？？？