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目前還找不到解釋的有"PP1-DS-175"

Ivy: 如果兩者的SUM為負數時,則無法成立 ex: S{SUM}= -35 = T{SUM}  

                                                                S{AVG}=(-35)／5= -7

                                                                T{AVG}=(-35)／7= -5     

                                                           此時, S{AVG} < T{AVG} 

                                                           BUT  S{個數=5}  < T {個數=7} 

                                                      -->跟SUM為正時得到的結果相反!!

大家一起幫忙想想看吧!!!

然後Alice問的"PP1-DS-159"

我上傷咖的討論精華去看人家講解

也是有人說直接帶數字進去看就好了

(1) n≠2k -->n=1,3,5,7,9 (odd)  :insufficient

(2) n≠3k -->n=1,2,4,5,7           :insufficient

(1)+(2) n≠2k 且 n≠3k --> n=1,5,7,...  代入(n-1)(n+1)

                                         n=1  (n-1)(n+1)=0 --> r=0

                                         n=5  (n-1)(n+1)=4*6=24--> r=0

                                         n=7  (n-1)(n+1)=6*8=24*2 --> r=0

                                    ---> (1)+(2) 充分

   另外我想到一個比較"通則"但麻煩的解法, 有興趣的人再看吧~

 (n-1) , n , (n+1) consecutive integers

(1)     n≠2k --> (n-1) =2k, 

                       (n+1)= 2k+2 =2(k+1)

                (n-1)*(n+1) = 2k*2(k+1) = 4*k*(k+1) 

                {if  k(k+1)=6t , r=0

            but if k(k+1)≠6t , r≠0}    --> (1)單獨：insufficient

(2)     n≠3k --> (n-1)=3Q

                    or (n+1)=3Q  {3個連續整數中必有一個為三的倍數}

           只能知道 (n-1)*(n+1) =3Q  --> (2)單獨：insufficient

(1)+(2)：(n-1)*(n+1) =3Q*4k(k+1)=12Q*k*(k+1) 

         --> when k= even{多了一個2} , (n-1)*(n+1)= 24T , r=0；

              when k= odd, (k+1)= even, (n-1)*(n+1) =24T, r=0

      　 故　(1)+(2): sufficient!

先這樣