FF MATH 2-43 AND 4-14

wawa0916 · 发表于 2006-2-17 01:28:00

2-43
TU兩位數, T是十位數,U是個位數,TU  is twice  product of these digits, U=?
答案是 6  可是我不知道怎摩算

4-14 N is the sum of the first consecutive intergers, where  101<=k<= 150, what is the sum of 1/n?
答案為 2*(1/101-1/105)

想請問一下  為什麼不能用(101+150)*50/2  的倒數  當答案呢???

hitlzc · 发表于 2006-2-17 09:07:00

2-43
TU兩位數, T是十位數,U是個位數,TU is twice product of these digits, U=?
答案是 6 可是我不知道怎摩算

10T+U=2*T*U

U=10T/(2T-1)=(10T-5+5)/(2T-1)=5+5/(2T-1)

因为T的范围是1-9的整数，令上式U为0-9整数的只有T=3

所以U=6

4-14 N is the sum of the first consecutive intergers, where 101<=k<= 150, what is the sum of 1/n?
答案為 2*(1/101-1/105)

想請問一下為什麼不能用(101+150)*50/2 的倒數當答案呢???

因为你那样理解题意不对。

wawa0916 · 发表于 2006-2-18 02:49:00

我了解了謝謝

jealv · 发表于 2006-6-10 13:59:00

4-14 N is the sum of the first consecutive intergers, where 101<=k<= 150, what is the sum of 1/n?
答案為 2*(1/101-1/105)

题意怎么理解呢？原题就是这样写的吗？
多谢~~~

yaoyao99 · 发表于 2006-6-11 06:38:00

http://forum.chasedream.com/dispbbs.asp?boardID=22&ID=176240&page=1

FF MATH 2-43 AND 4-14

FF MATH 2-43 AND 4-14

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