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教育王國 討論區 小學雜談 請教數學問題,請多多幫忙!
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請教數學問題,請多多幫忙!

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3315
發表於 13-3-25 13:48 |顯示全部帖子

我小四兒的數學興趣班中,有一條數我不懂,請各位幫忙。
請教我甚樣計算。

Question:
Here is an ancient Vedic method of testing for divisibility by 7.

A number is divisible by 7 if and only if the same is true of the number formed by adding 5 times the last digit to the number with the last digit removed.

The test is continued until a number is reached that is small enought to check easily.
To test 686, for example we added 5 x6 =30 to 68, giving 98.
Then add 5 x8 =40 to 9, giving 49. Since 7 divides 49, it also divides 686.
To test 689, we add 5 x9 =45 to 68, givng 113.
Then add 5 x3 = 15 to 11, giving 26. Since 7 does not divide 26, it does not divide 689.

Proof  & reason of listed;

Use the number 2534 to explain WHY the Vedic method works.


To test 2534, for example 5 x4=20 to 253, giving 273
Then add 5 x3 =15 to 27, giving 42, Since 7 divides 42. It also divide 2534.

I can calcule but can not explaination?

Rank: 6Rank: 6


6160
發表於 13-3-25 14:23 |顯示全部帖子
Let the number be XXXXa and the remainder of dividing XXXX by 7 be b.

We have to show 10b + a is divisible by 7 if and only if b + 5a is.

If 10b + a = 0 (mod 7), then b + 5a = 7(3b + a) - 2(10b + a) = 0 (mod 7).

If b + 5a = 0 (mod 7), then 10b + a = 10(b + 5a) - 49a = 0 (mod 7).

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3315
發表於 13-3-25 15:24 |顯示全部帖子
A more complicated algorithm for testing divisibility by 7 uses the fact that 100 ≡ 1, 101 ≡ 3, 102 ≡ 2, 103 ≡ 6, 104 ≡ 4, 105 ≡ 5, 106 ≡ 1, ... (mod 7). Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if the number obtained using this procedure is divisible by 7 (hence 371 is divisible by 7 since 28 is).[8]

This method can be simplified by removing the need to multiply. All it would take with this simplification is to memorize the sequence above (132645...), and to add and subtract, but always working with one-digit numbers.

點評

Unclejt  How come 100 ≡ 1, 102 ≡ 2, 103 ≡ 6, 104 ≡ 4, 105 ≡ 5 (mod 7)?  發表於 13-3-25 16:42

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3315
發表於 13-3-25 15:26 |顯示全部帖子
This is other method, but how to proof that?
My son can calculate but can not proof.

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3315
發表於 13-3-25 15:30 |顯示全部帖子
Unclejt 發表於 13-3-25 14:23
Let the number be XXXXa and the remainder of dividing XXXX by 7 be b.

We have to show 10b + a is di ...

Thank you.

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8834
發表於 13-3-25 18:59 |顯示全部帖子
本帖最後由 JustAParent 於 13-3-25 19:15 編輯
sampapa 發表於 13-3-25 15:24
A more complicated algorithm for testing divisibility by 7 uses the fact that 100 ≡ 1, 101 ≡ 3, 10 ...

對讀過number theory的人來說, 其實這個方法一點也不複雜.  但上述方法既無使用7這個數的特異性, 也不快速, 用來玩玩基礎number theory概念可以, 無實際用途的.

假設一個6位數為N = abcdef,

N = abcdef = a*10^5 + b*10^4 + c*10^3 + d*10^2 + e*10^1 + f*10^0
                  = a*(14285*7+5) + b*(1428*7+4) + c*(142*7+6) + d*(14*7+2) + e*(1*7+3) + f*(1)
                  = a*(14285*7) + b*(1428*7) + c*(142*7) + d*(14*7+2) + e*(1*7) + f*(1) + a*5 + b*4 + c*6 + d*2 + e*3 + f*1

因此, 驗證abcdef可否被7整除只須驗證a*5 + b*4 + c*6 + d*2 + e*3 + f*1是否7的倍數.

同理地, 我們可把証明延伸到任何多位數, 留給你小朋友試試吧.

其實這類貼根本不屬於小學雜談範疇, 我也懷疑有多少家長感興趣, 建議不要再貼這裡了.

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3315
發表於 13-3-25 20:20 |顯示全部帖子

引用:+本帖最後由+JustAParent+於+13-3-25+19:15

原帖由 JustAParent 於 13-03-25 發表
本帖最後由 JustAParent 於 13-3-25 19:15 編輯
謝謝你,我都知道這些數學題目對絶大部份的小學家長是沒有興趣的。但當小學生再升級的話,便要接觸這類型的題目。
由於我身邊沒有這方面的知識。只有在這兒請敎各高人。物理,化學及財務科我都有可應付及幫助。



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1524
發表於 13-3-28 17:18 |顯示全部帖子
本帖最後由 laorenjia 於 13-3-28 17:20 編輯

回復 sampapa 的帖子

After seeing what you said about your son, maybe it's worthwhile to write Unclejt's solution in a simpler way a primary school student can understand more easily.
Any integer can be written in the form of 10a+b.
example, the number 2534 above can be written as 253 x 10 + 4. If a+5b=7m, we multiply both sides by 10 to have 10a + 50b = 70m. Then we deduct 49b from both sides, we have 10a + b = 70m - 49b, which is clearly divisible by 7.

The student can then be asked to work out what other number(s)besides 5 can do the same trick.


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3315
發表於 13-3-28 17:54 |顯示全部帖子

引用:+本帖最後由+laorenjia+於+13-3-28+17:20+

原帖由 laorenjia 於 13-03-28 發表
本帖最後由 laorenjia 於 13-3-28 17:20 編輯

回復 sampapa 的帖子
很多謝你多次幫忙。我那問題兒童的怪問題。我皃都用Unclejt的方法作答。和解釋原因。