由 Raceleader 於 星期四 四月 03, 2003 10:15 am
問題1:787
問題2:116/9
問題3:(1,2,-3),(1,-3,2),(2,1,-3),(2,-3,1),(-3,1,2) or (-3,2,1)
問題4:250(√3+2)m
問題5:75
第一題:
Let the smallest number be k:
Since it has 6 types consecute sum, we can write as follows:
k=a+(a+1)=2a+1
k=b+(b+1)+(b+2)=3a+3
k=c+(c+1)+(c+2)+(c+3)=4c+6
k=d+(d+1)+(d+2)+(d+3)+(d+4)=5d+10
k=e+(e+1)+(e+2)+(e+3)+(e+4)+(e+5)=6e+15
k=f+(f+1)+(f+2)+(f+3)+(f+4)+(f+5)+(f+6)=7f+21
but 4c+6 must be even, and 2a+1 and 6e+15 must be odd, so we delete 4c+6, then add:
k=g+(g+1)+(g+2)+(g+3)+(g+4)+(g+5)+(g+6)+(g+7)=8g+28
this can also be deleted because it must be even, so we add:
k=h+(h+1)+(h+2)+(h+3)+(h+4)+(h+5)+(h+6)+(h+7)+(h+8)=9h+36
from 3a+3, 5d+10, 7f+21 and 9h+36, we know that k is the multiple of 315.
We found that when k=315, then (a,b,d,e,f,h)=(157,104,61,50,42,31)
So, the smallest number is 315