[計量] PP1-Q7

看板GMAT (GMAT入學考試)作者 (康)時間16年前 (2008/07/23 11:30), 編輯推噓2(202)
留言4則, 3人參與, 最新討論串1/1
7. For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is (A) between 2 and 10 (B) between 10 and 20 (C) between 20 and 30 (D) between 30 and 40 (E) greater than 40 這題解法由(2*4*6*8...*100)+1=2^50*50!+1 解起 我想問的是為什麼(2*4*6*8...*100)+1 會等於 2^50*50!+1 ? -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 163.13.36.176

07/23 11:51, , 1F
(2*4*6*8...*100)+1 = 2[1*2*3*4....*50]+1
07/23 11:51, 1F

07/23 12:35, , 2F
所以是 2*50!+1而不是 2^50*50!+1 ?
07/23 12:35, 2F

07/23 20:19, , 3F
2要提五十次出來~所以是2^50
07/23 20:19, 3F

07/23 20:19, , 4F
請問接下來怎麼解啊?
07/23 20:19, 4F
文章代碼(AID): #18XgMuDw (GMAT)
文章代碼(AID): #18XgMuDw (GMAT)