jj zhong jie --help!

andrewcan

I doubt the original keys from jj, anyone please give instruction: THanKs!

171.set S 有240项，问第239项是多少
(1)除第一项外，每项都比前一项少4
(2)239项和第一项差952


168.X=K^4 ，K是正整数，X被32除后余数是0。问K被32整除，余数可能是：（1）2，（2）4，（3）6




130.一个数列，A1=2，A2=－3，A3=5，A4=－1，以后An=An－4，问数列的前97项和是多少？

mihu

168. Key is (2)

X = 32m = 2^4 * 2m = 2^4 * 2^4 * n^4 = K^4
so K = 4*n
since K and 32 both have factor 4, so 余数 must have a factor 4
choose (2)

note: 1. m is a whole number
      2. since X = K^4, so 2m must be a form of 2^4 * n^4, n is a whole number

what is JJ's key?

[此贴子已经被作者于2002-11-15 5:02:58编辑过]

andrewcan

many thanks, Mihu.

your key to 168 agrees with the jj key: 4.

the other two: for 130:74; for 171: D.
I still do not quite understand the above two.

tongxun

171、E。因为没有具体的数值可供计算。
130、题干应该是An=A（n－4）。
所有的数字，每4个循环一次。97/4 =24 余 1，
每4项相加=5，则24*5=120，
余1，应为2，
所以和=120+2=122

andrewcan

thanks a lot Tongxun!

mihu

* a shortcut to 168: assume X=64, then K=4, so 4 mod 32 = 4, choose (2).