小学超级奥数题目

来源：百度知道 编辑：UC知道 时间：2024/09/22 05:29:12
1. 对于每个正整数n， 让f(n)代表最小的正整数s, 并且1+2+3+...+(s-1) +s 的值能被n整除。举个例子，f(5) = 4因为1+2+3+4的值能够被5整除，而1或1+2或1+2+3 都不能被5整除。
a) 找出所有正整数a的值，使f(a)= 8
b) 证明满足 f(b+1) - f(b) > 2009 的正整数b有无数个
c) 找出并证明能够使f(c) = f(c+k)中c有正奇数解的k的最小正整数值。

2. a,b,c 都是正整数，找出所有组（a,b,c) ， 使 a!=4(b!) + 10 (c!)

a) 1+2+3+...+(s-1) +s= s(s+1)/2 = 36
Consider all the divisors of 36, and notice it is impossible for n to be any odd number less than 9, since simply let s=n<9 then n|s(s+1)/2 holds, thus s<8, contradiction.
so we obtain,
f(4)=7, f(6)=3, f(9)=8, f(12)=8, f(18)=8, f(36)=8

all the possible values of a are 9,12,18,36.

b) Lemma 1: if m is an odd prime, f(m)<=m-1
Lemma 2: if n is a power of 2, namely, n=2^k, f(n)=2n-1

Proof of Lemma 2 is obvious.
Pf of Lemma 1: m|s(s+1)/2, m is odd, so m|s(s+1), s=m-1 satisfies.
Thus s<=m-1

Proof of (b): Let b=2^k-1, f(b+1)-f(b)>= 2b+1 - (b-1)=b+2
if b >2007, f(b+1) - f(b) > 2009
and there exists infinitely many b=2^k-1>2007.
Thus completes the proof.

(c) Prove: f(c)=f(c+1) and f(c)=f(c+2) have no odd solutions.
Pf: (c,c+1)=1
if f(c)=f(c+1)=s, then c|s(s+1)/2, and c+1|s(s+1)/2
thus, c(c+1)|s(s+1)/2
c(c+1)<