Re: 帮个忙否?

楼主: hiei81 (宝贝。永远)   2004-11-03 13:07:49
※ 引述《chaogold (H.Y. Chao)》之铭言:
: 1.Find all integers n>=1 such that
: (n^3+3)/(n^2+7)is an intege
(n^2+7)|(n^3+7n) => (n^2+7)|(7n-3)
=> n<=5 => n=2,5
: 2.Find all integral values of x,y,z,w given
: that x!+y!=2^z*3^w
Consider x>0, y>0
WLOG, assume x>=y
Note y<5 otherwise 5|2^z+3^w
if y=1 => x=1 or 2 since if x>=3 then (2,x!+1)=(3,x!+1)=1
=> (1,1,1,0) or (2,1,0,1) (reverse x,y => (1,2,0,1))
if y=2 => if x=2 or 3 then z>1 if x>=4 then z=1 => 3|(x!/2+1) -><-
=> (2,2,2,0) or (3,2,3,0) (2,3,3,0)
if y=3 => if x>=6 then w=1 => 2|(x!/6+1) -><- so x<=5
=> (3,3,2,1)
if y=4 => if x>=6 then z=3 => 3|(x!/24+1)-><- so x<=5
=> (4,4,4,1) or (5,4,4,2) (4,5,4,2)
共10组解
: 3.Let f (x) be a linear function such that
: f(0)=–5 and f(f(0)) =–15.
Let f(x)=ax+b => b=-5, a=2
: Find all values of m for which the solutions
: of the inequality f(x)f(m–x)> 0 form an
: interval of length 2.
=>(2x-5)(-2x+2m-5)>0 => (x-5/2)(x-(5/2 -m))<0
=>5/2-m = 5/2 +- 2 => m=2,-2 (好怪的题目?)
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