5、对怎样的正整数n，集合 可以划分成5个两两不交的子集，使得每个子集的元素和相等。
1。
若正整数n满足上面的条件,则
5|1+2+。。+n=n(n+1)/2
==>
5|n或5|n+1
下面证明5|n或5|n+1,9≤n,满足上面的条件。
2。
5|n,10≤n
ⅰ。
n=10k,1≤k
St={t+10(s-1),10+1-t+10(s-1),1≤s≤k,1≤t≤5}
St中的数之和=55+(k-1)k,满足你的条件
ⅱ。
n=10k+15,0≤k
S1={15+1+10s,15+10+10s,0≤s≤k-1}∪{9,15}
S2={15+2+10s,15+9+10s,0≤s≤k-1}∪{10,14}
S3={15+3+10s,15+8+10s,0≤s≤k-1}∪...全部
1。
若正整数n满足上面的条件,则
5|1+2+。。+n=n(n+1)/2
==>
5|n或5|n+1
下面证明5|n或5|n+1,9≤n,满足上面的条件。
2。
5|n,10≤n
ⅰ。
n=10k,1≤k
St={t+10(s-1),10+1-t+10(s-1),1≤s≤k,1≤t≤5}
St中的数之和=55+(k-1)k,满足你的条件
ⅱ。
n=10k+15,0≤k
S1={15+1+10s,15+10+10s,0≤s≤k-1}∪{9,15}
S2={15+2+10s,15+9+10s,0≤s≤k-1}∪{10,14}
S3={15+3+10s,15+8+10s,0≤s≤k-1}∪{11,13}
S4={15+4+10s,15+7+10s,0≤s≤k-1}∪{4,8,12}
S5={15+5+10s,15+6+10s,0≤s≤k-1}∪{1,2,3,5,6,7}
St中的数之和满足你的条件。
3。
5|n+1,9≤n
ⅰ。
n=10k+9,0≤k
S1={9+1+10s,9+10+10s,0≤s≤k-1}∪{9}
S2={9+2+10s,9+9+10s,0≤s≤k-1}∪{1,8}
S3={9+3+10s,9+8+10s,0≤s≤k-1}∪{2,7}
S4={9+4+10s,9+7+10s,0≤s≤k-1}∪{3,6}
S5={9+5+10s,9+6+10s,0≤s≤k-1}∪{4,5}
St中的数之和满足你的条件。
ⅱ。
n=10k+14,0≤k
S1={14+1+10s,14+10+10s,0≤s≤k-1}∪{7,14}
S2={14+2+10s,14+9+10s,0≤s≤k-1}∪{8,13}
S3={14+3+10s,14+8+10s,0≤s≤k-1}∪{9,12}
S4={14+4+10s,14+7+10s,0≤s≤k-1}∪{10,11}
S5={14+5+10s,14+6+10s,0≤s≤k-1}∪{1,2,3,4,5,6}
St中的数之和满足你的条件。
所以满足你的条件的n=5k,5k-1,k>1。
。收起
