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有關帖文題目顯示香港字問題

一直以來,不少會員反映討論區帖文題目未能正常顯示香港字,由於此項更新牽涉整個系統,故一直未能完善。然而我們明白各會員對此有一定需求,決定於星期三早上7時短暫「熄登」更新系統,解決題目顯示問題,預計需時兩小時。不便之處,敬請原諒。

更新完成後題目將支援香港字、大部分日文及韓文字體,字數限制將由現時25個全形字符增加至30個。與此同時,討論區桌面版將增設M版現有的回帶及追蹤功能,發表頁面的題目輸入位置亦會加入字數提示功能,希望能改善大家的使用體驗。

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超簡單自創數學題
112 個回應
Find the number of ways that 11 identical balls can put into five different boxes.


Find the number of ways that 11 identical balls can put into five different boxes.

#hoho#cl


Find the number of ways that 11 identical balls can put into five different boxes.

#hoho#cl

A box may contain zero balls.


睇到呢抽數字睇到暈


Ax^2+Bx+C+D=0
Ax^2+Bx+B=0
So B/A=a+b and ab
So a^4+b^4=(a^2+b^2)^2-2(ab)^2=((B/A)^2-2(B/A))^2-2(B/A)^2
=(B/A)^4-4(B/A)^3+2(B/A)^2


x, y, z are integer that are side of an obtuse-angled triangle. If xy = 4, find z.


為左推動學術台發展, 我決定答岩有一膠幣以示鼓勵


Find the number of ways that 11 identical balls can put into five different boxes.

#hoho#cl

A box may contain zero balls.


I don’t have a paper with me now.
An useful hint might be to consider a simpler problem, where you find the number of ways n balls can be distributed among 2 boxes (which is n+1) and apply the result recursively to generalise to 3 boxes and so on.


Given that f(0) =/= 0, find all function(s) such that f(x)f(y)=f(xy) for any real numbers x,y.#bye#

f(x) = 1/x
f(y) = 1/y
f(xy) = 1/xy = f(x)f(y)
搵到一個答案 #yup#


應該唔得,因為0嘅時候係undefined
不過如果幫0加exception嘅話,咁就無問題

無錯, 因為f(0)應該undefined.


應該係話你define唔到f(0)使得上述所有條件成立


x, y, z are integer that are side of an obtuse-angled triangle. If xy = 4, find z.


If x and y are both 2, then z=3 would have satisfied the requirement.
(Being greater than 2*sqrt(2))

If x=1 and y=4 or y=1 and x=4
z must be greater than 3 but less than 5
that is 4

But triangle with two sides of length 4 and the remaining side of length 1 is not an obtuse angled triangle.


Find the number of ways that 11 identical balls can put into five different boxes.

#hoho#cl

A box may contain zero balls.


I don’t have a paper with me now.
An useful hint might be to consider a simpler problem, where you find the number of ways n balls can be distributed among 2 boxes (which is n+1) and apply the result recursively to generalise to 3 boxes and so on.

I can write a formula without any steps. Think about it yourself.


x, y, z are integer that are side of an obtuse-angled triangle. If xy = 4, find z.


If x and y are both 2, then z=3 would have satisfied the requirement.
(Being greater than 2*sqrt(2))

If x=1 and y=4 or y=1 and x=4
z must be greater than 3 but less than 5
that is 4

But triangle with two sides of length 4 and the remaining side of length 1 is not an obtuse angled triangle.




Find the number of ways that 11 identical balls can put into five different boxes.


15C4=1365


Given that f(0) =/= 0, find all function(s) such that f(x)f(y)=f(xy) for any real numbers x,y.#bye#


let y = 0,
f(x) . f(0) = f(x . 0)
f(x) . f(0) = f(0)
f(x) . f(0) / f(0) = f(0) / f(0)
f(x) = 1

答案 f(x) = 1:)


Find all positive integers n such that (2^8+2^11+2^n) is a perfect square.


Find all positive integers n such that (2^8+2^11+2^n) is a perfect square.

2^8+2^11+2^n
=2^8+2(2)^4(2)^6+2^n
If n=12,
=2^8+2(2)^4(2)^6+2^12
=(2^4+2^6)^2
=80^2
.
.
.
uniqueness 有d難prove, 你等等
Find all positive integers n such that (2^8+2^11+2^n) is a perfect square.

2^8+2^11+2^n
=2^8+2(2)^4(2)^6+2^n
If n=12,
=2^8+2(2)^4(2)^6+2^12
=(2^4+2^6)^2
=80^2
.
.
.
uniqueness 有d難prove, 你等等。


Find all positive integers n such that (2^8+2^11+2^n) is a perfect square.

2^8+2^11+2^n
=2^8+2(2)^4(2)^6+2^n
If n=12,
=2^8+2(2)^4(2)^6+2^12
=(2^4+2^6)^2
=80^2
.
.
.
uniqueness 有d難prove, 你等等。


I had proven when n=12, 2^8+2^11+2^n is a perfect square.
I won't consider n<12 as it's easy to compute.
Now for n>12.


Find all positive integers n such that (2^8+2^11+2^n) is a perfect square.


2^8+2^11+2^n=2^8(1+8+2^(n-8))=2^8(9+2^(n-8))
n=12 is a solution, since 9+2^4=25 which is a perfect square.

To claim n=12 is the only solution, let n-8=N
and we require (3^2+2^N) to be perfect square, say a^2
then 2^N=(a+3)(a-3), which follows (a+3)=2^N1, (a-3)=2^N2
implying 2^N1 is 6 greater than 2^N2

But the powers of 2 are 2, 4, 8, 16, 32, 64, 128, ...
and it can been seen that N1=3, N2=1 is the only possible combination

thus proving uniqueness.


Find the number of ways that 11 identical balls can put into five different boxes.

#hoho#cl

A box may contain zero balls.


I don’t have a paper with me now.
An useful hint might be to consider a simpler problem, where you find the number of ways n balls can be distributed among 2 boxes (which is n+1) and apply the result recursively to generalise to 3 boxes and so on.

我都係咁諗不過條式會好長 唔想計 xx(


Find all positive integers n such that (2^8+2^11+2^n) is a perfect square.


2^8+2^11+2^n=2^8(1+8+2^(n-8))=2^8(9+2^(n-8))
n=12 is a solution, since 9+2^4=25 which is a perfect square.

To claim n=12 is the only solution, let n-8=N
and we require (3^2+2^N) to be perfect square, say a^2
then 2^N=(a+3)(a-3), which follows (a+3)=2^N1, (a-3)=2^N2
implying 2^N1 is 6 greater than 2^N2

But the powers of 2 are 2, 4, 8, 16, 32, 64, 128, ...
and it can been seen that N1=3, N2=1 is the only possible combination

thus proving uniqueness.

其實最後部份
因為6=2×3
所以直接寫2^N2=2及2^(N1-N2)-1=3便足夠


Find all polynomial f(x) such that f(x^2)=x^3 f(x+1)-2x^4+2x^2 for any real value of x.


Find all polynomial f(x) such that f(x^2)=x^3 f(x+1)-2x^4+2x^2 for any real value of x.

目前搵到一條:f(x)=x^3-2x^2+2x


Find all polynomial f(x) such that f(x^2)=x^3 f(x+1)-2x^4+2x^2 for any real value of x.

目前搵到一條:f(x)=x^3-2x^2+2x

Step: f(0)=0^3f(1)-2(0)^4+2(0)^2=0
f((-1)^2)=(-1)^3f(0)-2(-1)^4+2(-1)^2
f(1)=0
f(1)=(1)f(2)-2+2=f(2)
hence x, (x-1), (x-2) are factors of f(x)
This means degree of f(x) >=3
Suppose that degree of f(x) is n, from f(x^2)=x^3 f(x+1)-2x^4+2x^2---*
degree of L.H.S. of (*)= 2n
degree of R.H.S. of (*)= n+3
L.H.S.=R.H.S. implies 2n=n+3, n=3
Since x, (x-1), (x-2) are three factors of f(x) (proven),
Let f(x)=Cx(x-1)(x-2)= C(x^3-2x^2+2x) where C is constant
By(*),
C(x^6-3x^4+2x^2)=Cx^3((x+1)^3-3(x+1)^2+2(x+1))-2x^4+2x^2
C(x^6-3x^4+2x^2)=Cx^3(x^3-x)-2x^4+2x^2
C(x^6-3x^4+2x^2)=Cx^6-(2+C)x^4+2x^2
Hence -3C=-2-C and 2C=2 gives C=1,
f(x)=x^3-2x^2+2x Q.E.D.


Find all polynomial f(x) such that f(x^2)=x^3 f(x+1)-2x^4+2x^2 for any real value of x.

目前搵到一條:f(x)=x^3-2x^2+2x

Step: f(0)=0^3f(1)-2(0)^4+2(0)^2=0
f((-1)^2)=(-1)^3f(0)-2(-1)^4+2(-1)^2
f(1)=0
f(1)=(1)f(2)-2+2=f(2)
hence x, (x-1), (x-2) are factors of f(x)
This means degree of f(x) >=3
Suppose that degree of f(x) is n, from f(x^2)=x^3 f(x+1)-2x^4+2x^2---*
degree of L.H.S. of (*)= 2n
degree of R.H.S. of (*)= n+3
L.H.S.=R.H.S. implies 2n=n+3, n=3
Since x, (x-1), (x-2) are three factors of f(x) (proven),
Let f(x)=Cx(x-1)(x-2)= C(x^3-3x^2+2x) where C is constant
By(*),
C(x^6-3x^4+2x^2)=Cx^3((x+1)^3-3(x+1)^2+2(x+1))-2x^4+2x^2
C(x^6-3x^4+2x^2)=Cx^3(x^3-x)-2x^4+2x^2
C(x^6-3x^4+2x^2)=Cx^6-(2+C)x^4+2x^2
Hence -3C=-2-C and 2C=2 gives C=1,
f(x)=x^3-3x^2+2x Q.E.D.



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