请教Prep2# 数学^.^

Carrie2009

1、If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of aⁿ, what is the value of a ?
(1)aⁿ = 64
(2)n = 6

这道题答案是B。如下：(1)
2^6= 64 , 4^3=64 , 8^2=64 , not sufficient
(2) n = 6 , 2^6= 64 , sufficient
但是为什么能根据n=6 就确定a=2呢？

2、If x and y are positive, which of the following must be greater than  1/√(x+y)?
    1. √(x+y)/2x
    2. (√x +√y)/(x+y)
    3. (√x -√y)/(x+y)
这道题的答案是2only。
如果是用不等式来解的话可以得出答案，但是非常耗时。大家有没有什么快速点的方法呢？

PS. 这个东东"√"是根号。

danoulini

product of the first 8 positive integers is a multiple of a^n:

1*2*...*8 = 2^7 * 3^2 * 5 * 7 = k * a^n

中间的那个式子不可能在变换成某一个比2大的数的6次方作因子的式子了，2是极限

所以a=2

Carrie2009

合并合并。。。我怎么就没想到呢，多谢多谢~^.^

Carrie2009

第二题有什么好方法吗？

Carrie2009

up~~

Carrie2009

sdcar2010

2、If x and y are positive, which of the following must be greater than  1/√(x+y)?
   1. √(x+y)/2x
   2. (√x +√y)/(x+y)
   3. (√x -√y)/(x+y)

1. Only works if (x+y) > 2x or y > x
So the answer is "it depends."

2. Compare the squre of both values. (x + 2√(xy) + y)/(x +y)^2 vs. 1/(x + y)
Since (x + 2√(xy) + y) is greater than (x +y), (√x +√y)/(x+y) is greater than 1/√(x+y)
So the answer is yes.

3. Compre the squre of both values again. (x - 2√(xy) + y)/(x +y)^2 vs. 1/(x + y)
Since (x - 2√(xy) + y) is smaller than (x +y), (√x - √y)/(x+y) is smaller than 1/√(x+y)
So the answer is never.

Carrie2009

Thank you lovely~