某一條數學假說及它的推廣

說出來也為羞愧的,因為我自中四第一次見到此假說,過了超過十年竟然想不出如何用嚴僅的推理來證明,而它也是超級複雜的東西,它只有兩條演算規律:

A. 如果是單數則加1,
B. 如果是雙數則除2;
要證明的是任何一個正整數,在經過有限的步驟後最終一定會陷入1-2-1的死迴圈;

我想來想去的結論都是: 如何證明任何一個正整數都可以用2的升冪來表示呢?
即任何數=n(1)*2^0+n(2)*2^1+n(3)*2^2+.....n(k)*2^(k-1)

只要能證明此前設,其實即2進制可以成立的基礎,則不難證明任何一個單數:
A. 一定是先經A步再行B步,得出比本身少的數值,如25,26,13,14,7,8,4,2,1,因此這是遞減數列,數值只會愈來愈細,不會愈來愈大,因為數值本身是有限,所以最終一定在經過有限的步驟後最終一定會陷入1-2-1的死迴圈;

如果是任何一個雙數,同理會形成遞減數列,數值只會愈來愈細,不會愈來愈大,再分兩類:
A. 2的完整乘冪,如2的7次方128,則會直降到1,最終一定會陷入1-2-1的死迴圈,步驟數為2的完整次方數;
B. 2的不完整乘冪,如2的7次方加6次方192,則會經歷一有限次數的B步而成為單數,在此例中則是經歷6次B步後成為3,因為是單數,所以可以應用上面單數的證明來解決。

算不算證明完畢?

其實有更有趣的事,我想問的是到底在十進制可以有幾多條如此的假說,是不是有限數?而每一個假說又有什麼特色?
例如用三條演算規律:
A. 如果用3除剩1時是加2,
B. 如果用3除剩2時是加1,
C. 如果用3可整除時是除3,
要證明的是任何一個正整數,在經過有限的步驟後最終一定會陷入3-1-3的死迴圈;

單是加不夠好玩,不如改成:
例如用三條演算規律:
A. 如果用3除剩1時是減2,
B. 如果用3除剩2時是加1,
C. 如果用3可整除時是除3,
要證明的是任何一個正整數,在經過有限的步驟後最終一定會陷入3-1-3的死迴圈;

留意到什麼沒有?
理論上,我可以隨便作一條出來:
A. 如果用5除剩1時是減2,
B. 如果用5除剩2時是加1,
C. 如果用5可剩3時是減3,
D. 如果用5除剩4時是加1,
E. 如果用5可整除時是除3,
要證明的是任何一個正整數,在經過有限的步驟後最終一定會陷入5-1的死迴圈。

(未完待續....)

Two corollaries of ‘Divisibility of 11'

We notice that in decimal (10-base system), we can evaluate the divisibility of 9 by adding up all its digits and test the divisibility of the sum of the digits. While also in decimal system, we can evaluate the divisibility of 11 by adding up its odd digits and even digits and test the divisibility of the differences of the sum of the digits.

Following the logic in the article that discussing the divisibility of 9, we can made the following corollary: Similar to the idea that we can test test the divisibility of the differences of the sum of the digits, we could also test the divisibility of 12 by the differences of the sum of the odd and even digits in a base 11 system, the divisibility of 13 by the differences of the sum of the odd and even digits in a base 12 system…. etc.

Moreover, I made the hypothesis that in decimal system, divisibility by 12 could be tested by the divisibility of the weighted differences of the sum of the digits, i.e. even digits sum*2- odd digits sum. Moreover, divisibility by 13 could be tested by the divisibility of the weighted differences of the sum of the digits, i.e. even digits sum*3- odd digits sum, divisibility by 14 could be tested by the divisibility of the weighted differences of the sum of the digits, i.e. even digits sum*4- odd digits sum…. etc.
And in general, divisibility by n+x in a n-base system could be tested by the divisibility of the weighted differences of the sum of the digits, i.e. even digits sum*n- odd digits sum(in the same system.)

Modulus of 9

We all learn from our school that to test the divisibility of a number by 9, we just need to add up their digits and see if the sum of digits is divisible by 9. It is actually beyond that, the remain of a number divided by 9 is equal to the remain of the sum of digits by 9, which we can prove here:

Assume the digits are n(1)n(2)n(3)….n(k) in decimal system,
The value of number= n(1)*10^(k-1)+n(2)*10^(k-2)+n(3)*10^(k-3)+….+n(k)
= n(1)*[10^(k-1)-1]+n(2)*[10^(k-2)-1]+n(3)*[10^(k-3)-1}+....+[n(1)+n(2)+n(3)+....+n(k)]

Given that 10^n-1 is also divisible by 9 when n>=1,
thus the remain of the number divided by 9 is equal to the remain of the sum [n(1)+n(2)+n(3)+....+n(k)] divided by 9
which means the remain of the sum of its digits in decimal system when divided by 9 is equal to the remainder of the value of the number when divided by 9.

To give an example,
12345678=1*10^7+2*10^6+3*10^5+4*10^4+5*10^3+6*10^2+7*10^1+8
=(1*10^7+2*10^6+3*10^5+4*10^4+5*10^3+6*10^2+7*10^1+8 )
=((1*9999999+1)+(2*999999+2)+(3*99999+3)+(4*9999+4)+(5*999+5)+(6*99+6)+(7*9+7)+8 )
=(1*9999999+2*999999+3*99999+4*9999+5*999+6*99+7*9)+(1+2+3+4+5+6+7+8 )
(1*9999999+2*999999+3*99999+4*9999+5*999+6*99+7*9) is divisible by 9
thus 12345678 modulus 9=(1++2+3+4+5+6+7+8 ) modulus 9=0
12345678 is divisible by 9

There is nothing special about that, however, if we consider a number in k-base system,
i.e. The value of 12345678(k)=(1*k^7+2*k^6+3*k^5+4*k^4+5*k^3+6*k^2+7*k^1+8 )

Consider that 12345678(k)=(1*k^7+2*k^6+3*k^5+4*k^4+5*k^3+6*k^2+7*k^1+8 )
=[1*(k^7-1)+2*(k^6-1)+3*(k^5-1)+4*(k^4-1)+5*(k^3-1)+6*(k^2-1)+7*(k^1-1)]+(1+2+3+4+5+6+7+8 )

By the remainder theorem,
[1*(k^7-1)+2*(k^6-1)+3*(k^5-1)+4*(k^4-1)+5*(k^3-1)+6*(k^2-1)+7*(k^1-1)] is divisible by k-1,
i.e. when k=1 is substitute into the polynomail, it give the value of zero.

Thus the modulus of 12345678 in k-base system when divided by k-1 is equal to the modulus of (1+2+3+4+5+6+7+8 ) when divided by k-1

Therefore we can generalize from this result that in 10-base system, we can test the divisibility of a number by 9 by testing its divisibility of its sum of digits. We could also test the divisibility of a number in 9 base-system by 8 by testing its divisibility of its sum of digits when divided by 8. We could also test the divisibility of a number in 8 base-system by 7 by testing its divisibility of its sum of digits when divided by 6;…. etc.