[高中]兩題數學.. 求XYZ..

[高中]兩題數學.. 求XYZ..

快被報告弄到死的高中生.. 於 星期四 二月 02, 2012 8:57 pm


第一題 : xyz=x+y+z

第二題 : 10x+y=x^(y)+y^(x)+xy , 求正整數解

拜託了>__<..

快被報告弄到死的高中生..
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kungfan 於 星期日 三月 18, 2012 9:29 pm


第一題:不失一般性可設x>=y>=z,
當z=1時,原式:xy=x+y+1,
   xy-x-y+1=2,(x-1)(y-1)=2,
   x-1=2,y-1=1,x=3,y=2
當z>=2時,
xyz>=xz^2 >= 4x
x+y+z<=3x 無解。
通解:(x,y,z)=(3,2,1)or(3,1,2)or(2,1,3)or(2,3,1)or(1,3,2)or(1,2,3)

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Re: [高中]兩題數學.. 求XYZ..

lskuo 於 星期一 三月 19, 2012 12:17 pm


快被報告弄到死的高中生.. 寫到:
第二題 : 10x+y=x^(y)+y^(x)+xy , 求正整數解

Case 1: If x = 1, then 10 + y = 1 + y + y, so y =9.

Case 2: If y = 1, then 10x + 1 = x + 1 + x, so x = 0. Not positive integer!!!

Next step is to find solutions other than 1 included.
Note that if there is any other positive-integer solutions, both x and y would be greater than 1.

Rewrite the equation as
10x + y - xy = 10x - y(x-1) = x^y + y^x

Note that the minimum of x^y + y^x would be 2^2 + 2^2 = 8. So

10x - y(x-1) = x^y + y^x > =  8

If x is known, then one can find the upper bound of y.

y < =  (10x-8)/(x-1) = 10 + 2/(x-1) < =  12

So 1 < y < =  12

Note that y = 12 occurs when x=2. But if (x,y) = (2,12)
10x - y(x-1) = 20 - 12  = 8
x^y + y^x =  2^12 + 12^2 = 4096 + 144

Therefore (x,y) = (2,12) is NOT the solution.

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Re: [高中]兩題數學.. 求XYZ..

lskuo 於 星期一 三月 19, 2012 12:19 pm


Continue (因為預覽時總是有些錯誤, 所以分段寫)

On the other hand, if y is known, then

x > =  (8-y)/(10-y) = 1 + 2/(y-10) > 1

So y > 10

Therefore, the only possible solution of y is 11.
If y = 11, then
10x + 11 = x^11 + 11^x + 11x
11 - x = x^11 + 11^x > x^11 + 11^1 = x^11 + 11
-x > x^11.  NO SUCH x (positive integer).

So the only solution is (x,y) = (1,9).

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kungfan 於 星期三 三月 21, 2012 4:04 pm


(x,y)=(2,3)也是一組解

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Re: [高中]兩題數學.. 求XYZ..

lskuo 於 星期三 三月 21, 2012 8:11 pm


lskuo 寫到:Continue (因為預覽時總是有些錯誤, 所以分段寫)

On the other hand, if y is known, then

x > =  (8-y)/(10-y) = 1 + 2/(y-10) > 1

So y > 10

Therefore, the only possible solution of y is 11.
If y = 11, then
10x + 11 = x^11 + 11^x + 11x
11 - x = x^11 + 11^x > x^11 + 11^1 = x^11 + 11
-x > x^11.  NO SUCH x (positive integer).

So the only solution is (x,y) = (1,9).


謝謝 kungfan 網友的指正.

錯誤原因出在方程式兩邊同除以負數時, 大小關係必須反向.
更正x,y均大於1的解法如下:

Because y(x-1) > 0,
10x - y(x-1) = x^y + y^x < 10x


x^y < 10x - x^y < = 10x
y ln x < ln 10 + ln x

(y-1) ln x < ln 10

ln x < ln 10 / (y-1)

2 < = x < 10^(1/(y-1))

log10(2)=0.3010 < 1/(y-1)
y < 1 + 1/log10(2) = 4.3


So now we know
  2 < = y < 4.3, and
  2 < = x < 10^(1/(y-1))

Try and error:
1. if y = 2, then x < 10
   10x + 2 = x^2 + 2^x + 2x
   x^2 = 8x + 2 - 2^x = even number
   NO positive-integer solutions (try x = 2,4,6,8)

2. if y = 3, then 2 < = x < 10^(1/2) < 3.17. Possible x are 2 and 3.
    10x + 3 - 3x = x^3 + 3^x
    x = 2

3. if y = 4, then x < 10^(1/3) < 2.16. So the only possible x is 2.
   10x + 4 - 4x = 16
   x^y + y^x = 2^4 + 4^2 = 32
   so (x,y) = (2,4) is not the solution.
  
  
The complete solutions are (x,y) = (1,9) and (2,3).

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devell 於 星期三 三月 21, 2012 11:30 pm


第一題  有要求一定要正整數解嗎?

根據算術平均數>= 幾何平均數  

( x+y+z ) / 3 >= ( xyz ) 的立方根

( x+y+z )^3 / 27 >= ( xyz )

( xyz )^3 /27 >= ( xyz )

( xyz )^2 >= 27

xyz >= 根號 27

等號成立時,x=y=z

則 x=y=z=根號3

這組是否也可以

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高中數學問題