教育王國

標題: F.2數學題 [打印本頁]

作者: cman_li 時間: 08-9-22 23:42 標題: F.2數學題

Suppose x,y,z is any integer

If 8x+4y-5z is divisible by 7, show that 9x+y+4z is also divisible by 7.

如果F.2計這條數會唔會太深

[ 本帖最後由 cman_li 於 08-9-23 09:35 編輯 ]

作者: swallowngxx 時間: 08-9-23 22:37

原帖由 cman_li 於 08-9-22 23:42 發表
Suppose x,y,z is any integer

If 8x+4y-5z is divisible by 7, show that 9x+y+4z is also divisible by 7.

如果F.2計這條數會唔會太深

x=y=z都是相同, 我仔仔剛上中二, 未學過這數式, 但他計到, 唔知算唔算深.

[ 本帖最後由 swallowngxx 於 08-9-23 22:39 編輯 ]

作者: cman_li 時間: 08-9-23 22:50

原帖由 swallowngxx 於 08-9-23 22:37 發表

x=y=z都是相同, 我仔仔剛上中二, 未學過這數式, 但他計到, 唔知算唔算深.

唔係計x,y,z 而係用 steps 去証明

Answer:

Let 8x+4y-5z = 7K where K is an integer

Firstly,

2 * ( 8x+ 4y - 5z ) - ( 9x + y + 4z )
=  16x + 8y -10z -9x -y -4z
= 7x +7y -14z
= 7 * (x+y-2z)
=7 * N                      where N is an integer

This implies that

2* (7K) - (9x+y+4z) = 7N

9+y+4z = 14K-7N
            = 7 *(2K-N)

Therefore 9x+y+4z is also divisible by 7

[ 本帖最後由 cman_li 於 08-9-23 23:02 編輯 ]

作者: daisychan 時間: 08-9-23 22:58

原帖由 swallowngxx 於 08-9-23 22:37 發表

x=y=z都是相同, 我仔仔剛上中二, 未學過這數式, 但他計到, 唔知算唔算深.

Dear swallowngxx,

I'm afraid your son is conceptually wrong.
He forgot the statement that " x, y, z are any integers"

8x + 4y - 5z is divisible by 7
Therefore,
16x + 8y - 10z is divisible by 7
16x + 8y - 10z - 7(x+y+z) is divisible by 7
i.e. 9x + y - 17z is divisible by 7
i.e. 9x + y - 17z +21z is divisible by 7
i.e. 9x +y + 4z is divisible by 7

I consider this question is difficult for most of the Form 2 students. Only the top 5% in Maths can tackle it.

[ 本帖最後由 daisychan 於 08-9-23 22:59 編輯 ]

作者: fungsir 時間: 08-9-23 23:16

原帖由 cman_li 於 08-9-22 23:42 發表
Suppose x,y,z is any integer

If 8x+4y-5z is divisible by 7, show that 9x+y+4z is also divisible by 7.

如果F.2計這條數會唔會太深

8x+4y-5z
= (7x+7y-7z)  +  (x-3y+2z) ....... (1)

∵ 8x+4y-5z  and  7x+7y-7z are divisible by 7

∴  x-3y+2z  is also divisible by 7 ...... (2)

9x+y+4z
= (8x+4y-5z)  + (x-3y+9z)
= (8x+4y-5z)  + (x-3y+2z)  + (7z)

∵ 8x+4y-5z  and  x-3y+2z  and  7z are divisible by 7

∴  9x+y+4z  is also divisible by 7

這題應不是中學的恆常課內容, 或許是奧數題吧.
未學過唔識不足為奇.  (即是深la)

作者: ZZdaphne 時間: 08-9-23 23:27

原帖由 fungsir 於 08-9-23 23:16 發表

8x+4y-5z
= (7x+7y-7z)  +  (x-3y+2z) ....... (1)

∵ 8x+4y-5z  and  7x+7y-7z are divisible by 7

∴  x-3y+2z  is also divisible by 7 ...... (2)

9x+y+4z
= (8x+4y-5z)  + (x-3y+9z)
= (8x+4y-5z)  + (x-3y+2z)  + (7z)

∵ 8x+4y-5z  and  x-3y+2z  and  7z are divisible by 7

∴  9x+y+4z  is also divisible by 7

這題應不是中學的恆常課內容, 或許是奧數題吧.
未學過唔識不足為奇.  (即是深la)

wo, cool

解釋好清楚，明白。
你好勁！！！！！
Fung sir

[ 本帖最後由 ZZdaphne 於 08-9-23 23:30 編輯 ]

作者: ZZdaphne 時間: 08-9-23 23:44

原帖由 daisychan 於 08-9-23 22:58 發表

Dear swallowngxx,

I'm afraid your son is conceptually wrong.
He forgot the statement that " x, y, z are any integers"

8x + 4y - 5z is divisible by 7
Therefore,
16x + 8y - 10z is divisible by 7
16 ...

呢個我都明！！！

作者: ZZdaphne 時間: 08-9-23 23:45

原帖由 cman_li 於 08-9-23 22:50 發表

唔係計x,y,z 而係用 steps 去証明

Answer:

Let 8x+4y-5z = 7K where K is an integer

Firstly,

2 * ( 8x+ 4y - 5z ) - ( 9x + y + 4z )
= 16x + 8y -10z -9x -y -4z
= 7x +7y -14z
= 7 * (x+y-2z)
=7 * N ...

呢個多左K同N，複雜左，唔明

作者: cman_li 時間: 08-9-24 08:55

原帖由 ZZdaphne 於 08-9-23 23:45 發表

呢個多左K同N，複雜左，唔明

整條數用左好多數學基礎知識

ii)  Remainder Theorem
iii) Operating with multivariable linear expression
-  Distributive law
-  Combine like term

The use of integer K and integer N is to apply Remainder Theorem, i.e.

Dividend = Divisor x Quotient + Remainder

In this question,

Dividend = 8x + 4y - 5z
Divisor = 7

The theorem state that 'Dividend' is divisible by 'Divisor' if and only if 'Remainder' is zero and 'Quotient' must be an integer.

Therefore we can assume Quotient = K and let

8x+4y-5z = 7 * K + zero
            = 7K

對於F.2學生,其實都教過以上知識,但同時運用在一條題目去表達就比較難(個人覺得)。其它兩位的解釋比較易明,但一樣運用以上知識。

P.S. 同阿仔一齊做數,人都年輕左,好似回到中學年代

。另外真係幾有滿足感,佢完全當我係偶像

[ 本帖最後由 cman_li 於 08-9-24 10:50 編輯 ]

作者: ZZdaphne 時間: 08-9-24 12:07

原帖由 cman_li 於 08-9-24 08:55 發表

P.S. 同阿仔一齊做數,人都年輕左,好似回到中學年代。另外真係幾有滿足感,佢完全當我係偶像

係呀，係呀

我個女F.1 ,
所以我未學到。

[ 本帖最後由 ZZdaphne 於 08-9-24 12:08 編輯 ]

作者: cman_li 時間: 08-9-24 12:58

原帖由 ZZdaphne 於 08-9-24 12:07 發表

係呀，係呀

我個女F.1 ,
所以我未學到。

語文佢話我out,數學就冇代溝。

有計傾。不過怕佢有數學恐懼。

[ 本帖最後由 cman_li 於 08-9-24 13:00 編輯 ]

歡迎光臨教育王國 (/)