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如何證明： 一個數值的所有數字加埋的數可以被3整除， 該數值也可被3整除。
最佳解答:
1+0+8 = 9 108除3 = 36 如此:207,306,405,504,603....801,900呢d 3位數字最易證明 其實： 個d數字係18中間加個0，27中間+個0...到81中間+個0 90中間+個0 ja 用係千位，萬位至到千億都得 總之就係一個能被3除盡的數中間+0 啦 2006-12-03 10:49:13 補充： 900一定可以，你可試90，將9-1 ， 0+1 ，就係81 如此類推...2-1 ， 7+1 就係18 1-1 8+ 1 就係9 啦 2006-12-06 13:55:37 補充： 899991都可以ga 3x3x3x3x11111 = 81 x11111因爲: 81 81 81 81 81 －－－－－－－ 899991 答案899991結論：即9倍數或3得倍數 x 11 或中間加 0 是3的倍數。 2006-12-06 14:03:15 補充： 如上，數字也可以是393 696 999 1899 2799 3699 4599 5499 6399 7299 8199 909918099 27099 36099 ...9099 18099 27099 36099的規則：9 18 27 36是9的倍數，在中間加0 後面加9（因爲0不能除任何數）393 696 999規則：393是36 x 11 696是63 x 11 999是9 x 1112799....的規則：18 27 36...是9的倍數 但中間不要加0 後面加9 。 2006-12-06 14:06:57 補充： 補充上面：333 666 999 是同一類，並不等於393 696的規則，還有的是2733 3633 這些都可以按照上面的規則説明。 2006-12-06 14:08:44 補充： The sum of the digits must be divisible by 3
其他解答:
12,15,18,21,24,27,30..........................3的倍數都可以，但一定要係兩位數以上。|||||In order to prove it, we need to prove this first, "10^n - 1 is always divisible by 3 for all positive integers n" This can be proven by mathematical induction. n=1, 10^n - 1 = 9 which is divisible by 3. Assume for n=k, 10^n - 1 is divisible by 3, i.e. 10^k - 1 = 3N where N is an integer n=k+1, 10^(k+1) - 1 = 10*10^k - 1 = 10^k - 1 + 9*10^k = 3N + 9*10^k = 3(N + 3*10^k) which is also divisible by 3. So by the principle of mathematical induction, 10^n - 1 is always divisible by 3 for all positive integers n. Now come to the question, any number can be expressed in a way like this, a_n*10^n + a_n-1*10^(n-1) + ... + a_2*10^2 + a_1*10 + a_0 where a_n, a_n-1, ... , a_2, a_1, a_0 are natural numbers in the range between 0 and 9 inclusively. Let f(n) represent the number, such that f(n) = a_n*10^n + a_n-1*10^(n-1) + ... + a_2*10^2 + a_1*10 + a_0 f(2) = a_2*100 + a_1*10 + a_0 = a_2*(99+1) + a_1*(9+1) + a_0 = a_2*99 + a_1*9 + (a_2 + a_1 + a_0) f(2) is divisible by 3 if (a_2 + a_1 + a_0) is divisible by 3. Similarly, f(3) = a_3*1000 + a_2*100 + a_1*10 + a_0 = a_3*(999+1) + a_2*99 + a_1*9 + (a_2 + a_1 + a_0) = a_3*999 + a_2*99 + a_1*9 + (a_3 + a_2 + a_1 + a_0) f(3) is divisible by 3 if (a_3 + a_2 + a_1 + a_0) is divisible by 3. Assume the following proposition is true for n=k, f(k) = a_k*10^k + a_k-1*10^(k-1) + ... + a_2*10^2 + a_1*10 + a_0 is divisible by 3 if (a_k + a_k-1 + ... + a_2 + a_1 + a_0) = 3M where M is an integer. f(k) = a_k*(10^k-1) + a_k-1*(10^(k-1)-1) + ... + a_2*99 + a_1*9 + (a_k + a_k-1 + ... + a_2 + a_1 + a_0) = a_k*(10^k-1) + a_k-1*(10^(k-1)-1) + ... + a_2*99 + a_1*9 + 3M f(k+1) = a_k+1*10^(k+1) + f(k) = a_k+1*(10^(k+1)-1) + f(k) + a_k+1 = (a_k+1*(10^(k+1)-1) + [a_k*(10^k-1) + a_k-1*(10^(k-1)-1) + ... + a_2*99 + a_1*9 + 3M] + a_k+1 = [(a_k+1*(10^(k+1)-1) + a_k*(10^k-1) + a_k-1*(10^(k-1)-1) + ... + a_2*99 + a_1*9] + (3M + a_k+1) = 3K + (3M + a_k+1) where K is an integer which was proven in the first part, "10^n - 1 is always divisible by 3 for all positive integers n". So f(k+1) is divisible by 3 if (3M + a_k+1) is divisible by 3. i.e. 3M + a_k+1 = 3P where P is another integer Note that 3M = a_k + a_k-1 + ... + a_2 + a_1 + a_0, therefore, 3M + a_k+1 = a_k+1 + a_k + a_k-1 + ... + a_2 + a_1 + a_0 = 3P Hence, f(k+1) is divisible by 3 if sum of their coefficients (a_k+1 + a_k + a_k-1 + ... + a_2 + a_1 + a_0) is divisible by 3. By the priniciple of mathematical induction, all any number f(n) to be divisible, the sum of its digits (a_0 + a_1 + ... + a_n), where natural numbers a_0, a_1, ..., a_n lie between 0 and 9 inclusively, has to be divisible by 3. 2006-12-03 12:08:10 補充： Interested readers may refer to the following link,http://en.wikipedia.org/wiki/Divisibility_rule#Divisibility_by_3 2006-12-03 12:51:08 補充： 如果只verify幾十個甚至頭幾千個, 頭幾萬個, 是不足夠證明"所有整數都成立"的.要證明"所有整數都適合", 可利用數學歸納法, 透過骨牌效應的原理, 來推斷"一個整數成立, 往後的整數皆成立". 2006-12-04 00:39:53 補充： 再重申, 只列出例子(12, 15, 18, 21, 27, ..., 十萬九千七加一, ...)不是證明"一個數值的所有數字加埋的數可以被3整除，該數值也可被3整除"的正確方法. 一定要用方法去證明所有, 是所有可能的出現的實數值中, 所有數字加埋的數可以被3整除, 該數值也可被3整除, 證明方為有效.一切列出的數字, 只能作為所有可能的出現的實數值的一部分, 能夠符合"一個數值的所有數字加埋的數可以被3整除，該數值也可被3整除"的命題, 而不能保證所有可能的出現的實數值都能符合. 這個是學數學很重要的邏輯概念. 2006-12-08 20:21:28 補充： 唔好意思, 今日再睇番, 又想再補充.見到以上的"393 696 999規則", "2799....的規則"等等, 都只係一些verification.要用此辦法, 就要首先找出所有可以被3整除的數所有出現的規則, 然後要歸納也好, 逐個逐個也好, 去證明該所有所有出現的規則, 都有"所有數字加埋的數可以被3整除, 該數值也可被3整除"的特性, 才能證明 in general 係成立的.|||||3的頭10個部數：3、6、9、12、15、18、21、24、27、30 所有數字和是：3、6、9、 3 、 6 、 9 、 3 、 6 、 9 、 3 因此可證明"一個數值的所有數字加埋的數可以被3整除"。|||||243=2+4+3=9 9可以比3整除 693=6+9+3=18 18可以比3整除 從以上的example把數字加埋,咁answer如果係3的倍數,咁就姐係可以比3整除。|||||冇得証明架喔~~你咪試下囉...FAD2A23AB937987B
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數學問題，數學問題，數學問題發問:
如何證明： 一個數值的所有數字加埋的數可以被3整除， 該數值也可被3整除。
最佳解答:
1+0+8 = 9 108除3 = 36 如此:207,306,405,504,603....801,900呢d 3位數字最易證明 其實： 個d數字係18中間加個0，27中間+個0...到81中間+個0 90中間+個0 ja 用係千位，萬位至到千億都得 總之就係一個能被3除盡的數中間+0 啦 2006-12-03 10:49:13 補充： 900一定可以，你可試90，將9-1 ， 0+1 ，就係81 如此類推...2-1 ， 7+1 就係18 1-1 8+ 1 就係9 啦 2006-12-06 13:55:37 補充： 899991都可以ga 3x3x3x3x11111 = 81 x11111因爲: 81 81 81 81 81 －－－－－－－ 899991 答案899991結論：即9倍數或3得倍數 x 11 或中間加 0 是3的倍數。 2006-12-06 14:03:15 補充： 如上，數字也可以是393 696 999 1899 2799 3699 4599 5499 6399 7299 8199 909918099 27099 36099 ...9099 18099 27099 36099的規則：9 18 27 36是9的倍數，在中間加0 後面加9（因爲0不能除任何數）393 696 999規則：393是36 x 11 696是63 x 11 999是9 x 1112799....的規則：18 27 36...是9的倍數 但中間不要加0 後面加9 。 2006-12-06 14:06:57 補充： 補充上面：333 666 999 是同一類，並不等於393 696的規則，還有的是2733 3633 這些都可以按照上面的規則説明。 2006-12-06 14:08:44 補充： The sum of the digits must be divisible by 3
其他解答:
12,15,18,21,24,27,30..........................3的倍數都可以，但一定要係兩位數以上。|||||In order to prove it, we need to prove this first, "10^n - 1 is always divisible by 3 for all positive integers n" This can be proven by mathematical induction. n=1, 10^n - 1 = 9 which is divisible by 3. Assume for n=k, 10^n - 1 is divisible by 3, i.e. 10^k - 1 = 3N where N is an integer n=k+1, 10^(k+1) - 1 = 10*10^k - 1 = 10^k - 1 + 9*10^k = 3N + 9*10^k = 3(N + 3*10^k) which is also divisible by 3. So by the principle of mathematical induction, 10^n - 1 is always divisible by 3 for all positive integers n. Now come to the question, any number can be expressed in a way like this, a_n*10^n + a_n-1*10^(n-1) + ... + a_2*10^2 + a_1*10 + a_0 where a_n, a_n-1, ... , a_2, a_1, a_0 are natural numbers in the range between 0 and 9 inclusively. Let f(n) represent the number, such that f(n) = a_n*10^n + a_n-1*10^(n-1) + ... + a_2*10^2 + a_1*10 + a_0 f(2) = a_2*100 + a_1*10 + a_0 = a_2*(99+1) + a_1*(9+1) + a_0 = a_2*99 + a_1*9 + (a_2 + a_1 + a_0) f(2) is divisible by 3 if (a_2 + a_1 + a_0) is divisible by 3. Similarly, f(3) = a_3*1000 + a_2*100 + a_1*10 + a_0 = a_3*(999+1) + a_2*99 + a_1*9 + (a_2 + a_1 + a_0) = a_3*999 + a_2*99 + a_1*9 + (a_3 + a_2 + a_1 + a_0) f(3) is divisible by 3 if (a_3 + a_2 + a_1 + a_0) is divisible by 3. Assume the following proposition is true for n=k, f(k) = a_k*10^k + a_k-1*10^(k-1) + ... + a_2*10^2 + a_1*10 + a_0 is divisible by 3 if (a_k + a_k-1 + ... + a_2 + a_1 + a_0) = 3M where M is an integer. f(k) = a_k*(10^k-1) + a_k-1*(10^(k-1)-1) + ... + a_2*99 + a_1*9 + (a_k + a_k-1 + ... + a_2 + a_1 + a_0) = a_k*(10^k-1) + a_k-1*(10^(k-1)-1) + ... + a_2*99 + a_1*9 + 3M f(k+1) = a_k+1*10^(k+1) + f(k) = a_k+1*(10^(k+1)-1) + f(k) + a_k+1 = (a_k+1*(10^(k+1)-1) + [a_k*(10^k-1) + a_k-1*(10^(k-1)-1) + ... + a_2*99 + a_1*9 + 3M] + a_k+1 = [(a_k+1*(10^(k+1)-1) + a_k*(10^k-1) + a_k-1*(10^(k-1)-1) + ... + a_2*99 + a_1*9] + (3M + a_k+1) = 3K + (3M + a_k+1) where K is an integer which was proven in the first part, "10^n - 1 is always divisible by 3 for all positive integers n". So f(k+1) is divisible by 3 if (3M + a_k+1) is divisible by 3. i.e. 3M + a_k+1 = 3P where P is another integer Note that 3M = a_k + a_k-1 + ... + a_2 + a_1 + a_0, therefore, 3M + a_k+1 = a_k+1 + a_k + a_k-1 + ... + a_2 + a_1 + a_0 = 3P Hence, f(k+1) is divisible by 3 if sum of their coefficients (a_k+1 + a_k + a_k-1 + ... + a_2 + a_1 + a_0) is divisible by 3. By the priniciple of mathematical induction, all any number f(n) to be divisible, the sum of its digits (a_0 + a_1 + ... + a_n), where natural numbers a_0, a_1, ..., a_n lie between 0 and 9 inclusively, has to be divisible by 3. 2006-12-03 12:08:10 補充： Interested readers may refer to the following link,http://en.wikipedia.org/wiki/Divisibility_rule#Divisibility_by_3 2006-12-03 12:51:08 補充： 如果只verify幾十個甚至頭幾千個, 頭幾萬個, 是不足夠證明"所有整數都成立"的.要證明"所有整數都適合", 可利用數學歸納法, 透過骨牌效應的原理, 來推斷"一個整數成立, 往後的整數皆成立". 2006-12-04 00:39:53 補充： 再重申, 只列出例子(12, 15, 18, 21, 27, ..., 十萬九千七加一, ...)不是證明"一個數值的所有數字加埋的數可以被3整除，該數值也可被3整除"的正確方法. 一定要用方法去證明所有, 是所有可能的出現的實數值中, 所有數字加埋的數可以被3整除, 該數值也可被3整除, 證明方為有效.一切列出的數字, 只能作為所有可能的出現的實數值的一部分, 能夠符合"一個數值的所有數字加埋的數可以被3整除，該數值也可被3整除"的命題, 而不能保證所有可能的出現的實數值都能符合. 這個是學數學很重要的邏輯概念. 2006-12-08 20:21:28 補充： 唔好意思, 今日再睇番, 又想再補充.見到以上的"393 696 999規則", "2799....的規則"等等, 都只係一些verification.要用此辦法, 就要首先找出所有可以被3整除的數所有出現的規則, 然後要歸納也好, 逐個逐個也好, 去證明該所有所有出現的規則, 都有"所有數字加埋的數可以被3整除, 該數值也可被3整除"的特性, 才能證明 in general 係成立的.|||||3的頭10個部數：3、6、9、12、15、18、21、24、27、30 所有數字和是：3、6、9、 3 、 6 、 9 、 3 、 6 、 9 、 3 因此可證明"一個數值的所有數字加埋的數可以被3整除"。|||||243=2+4+3=9 9可以比3整除 693=6+9+3=18 18可以比3整除 從以上的example把數字加埋,咁answer如果係3的倍數,咁就姐係可以比3整除。|||||冇得証明架喔~~你咪試下囉...FAD2A23AB937987B
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