[¼Æ¾Ç][°Q½×]13

[¼Æ¾Ç][°Q½×]13

¥Ñ E.T ©ó ¬P´Á¤é ¥|¤ë 06, 2003 6:45 pm


¥H¤T¦ì¼Æ¬°1³æ¦ì,±N©_¼Æ³æ¦ìªº¼Æ¦r©M¥[°_¨Ó»P°¸¼Æ³æ¦ìªº¬Û´î,¯à³Q13°£ºÉ´N¬O
¨Ò:5940129

940-(129+5)=806=13*62


why ??? how to proof this method ?
¡¸Click this : [>> ¦³½ìªº¬ì¾Ç(Science's World) <<]

Let's go to discuss ~*

¥ªÁä: ÂIÀ»ÁY©ñ; ¥kÁä: Æ[¬Ý­ì¹Ï¥ªÁä: ÂIÀ»ÁY©ñ; ¥kÁä: Æ[¬Ý­ì¹Ï¥ªÁä: ÂIÀ»ÁY©ñ; ¥kÁä: Æ[¬Ý­ì¹Ï¥ªÁä: ÂIÀ»ÁY©ñ; ¥kÁä: Æ[¬Ý­ì¹Ï

E.T

 
¤å³¹: 1486
µù¥U®É¶¡: 2003-03-29
¨Ó¦Û: ­»´ä

¥Ñ Raceleader ©ó ¬P´Á¤é ¥|¤ë 06, 2003 6:46 pm


You can prove this by mod

Raceleader
³X«È
 

¥Ñ kevin ©ó ¬P´Á¤é ¥|¤ë 06, 2003 6:49 pm


§Q¥Î7*11*13=1001=1000+1

kevin
±Ð¡@±Â
±Ð¡@±Â
 
¤å³¹: 1158
µù¥U®É¶¡: 2002-12-22

¥Ñ Raceleader ©ó ¬P´Á¤é ¥|¤ë 06, 2003 7:04 pm


Let a0, a1, a2,...,an are the 3-digit number:

The long number will be a0+103a1+106a2+...+103nan

106K-3+1=13A
106L-1=13B

Assume n is even:
if (a0+a2+...+an)-(a1+a3+...+an-1)=13N
then (a0+a2+...+an)-(a1+a3+...+an-1)+1001a1+999999a2+...+(103n-1)an=13N+13M
a0+101a1+102a2+...+10nan=13(M+N)

Assume n is odd:
if (a0+a2+...+an-1)-(a1+a3+...+an)=13N
then (a0+a2+...+an-1)-(a1+a3+...+an)+1001a1+999999a2+...+(103n+1)an=13N+13M
a0+101a1+102a2+...+10nan=13(M+N)

Raceleader
³X«È
 

¥Ñ Tassader-VIII ©ó ¬P´Á¤é ¥|¤ë 06, 2003 7:12 pm


¥Oabcdefghi¬O¾ã¼Æ
abcdefghi
=abc(10^6-1)+def(10^3+1)+[(abc)-(def)+(ghi)]

10^6-1=(10^2)^3-1=27-1=26=0(mod13)
10^3=1001=7*11*13=0(mod13)
§Y10^6-1»P10^3+1¬Ò¬°13¤§­¿¼Æ

¬G(abc)-(def)+(ghi)­Y¬°13¤§­¿¼Æ
¥iª¾abcdefghi¤]¬°13¤§­¿¼Æ

Tassader-VIII
³X«È
 

¥Ñ Tassader-VIII ©ó ¬P´Á¤é ¥|¤ë 06, 2003 7:13 pm


Raceleader³ºµM¤ñ§ÚÁÙ§Ö......... Ãø¹L¨ì­ú

Tassader-VIII
³X«È
 

¥Ñ Raceleader ©ó ¬P´Á¤é ¥|¤ë 06, 2003 7:14 pm


Don't be sad, 1000  £~£~£~

Raceleader
³X«È
 

¥Ñ Tassader-VIII ©ó ¬P´Á¤é ¥|¤ë 06, 2003 7:18 pm


Assume n is even:
if (a0+a2+...+an)-(a1+a3+...+an-1)=13N
then (a0+a2+...+an)-(a1+a3+...+an-1)+1001a1+999999a2+...+(103n-1)an=13N+13M
a0+101a1+102a2+...+10nan=13(M+N)

n is odd will have the similar result.
----------------------------
§A¼g³o¤@¦êÃÒ©ú¦³¥Î¶Ü???? §x´b


----------------------------------
Let a0, a1, a2,...,an are the 3-digit number:

The long number will be a0+103a1+106a2+...+103nan

106K-3+1=13A
106L-1=13B

------------------------------
Ãø¹D³o¼Ë¤£´N¨¬°÷»¡©ú¤F¶Ü????  §x´b

Tassader-VIII
³X«È
 

¥Ñ Raceleader ©ó ¬P´Á¤é ¥|¤ë 06, 2003 7:22 pm


103=-1(mod13)
So, 103(2n-1)=-1(mod13)
and 103(2n)=1(mod13)

Also, if n is odd, then put an to the odd side, nothing special

Raceleader
³X«È
 

¥Ñ Raceleader ©ó ¬P´Á¤é ¥|¤ë 06, 2003 7:25 pm


Tassader-VIII:
³o¼Ëº¡·N¤F§a  Åå¹Ä¸¹

Raceleader
³X«È
 

¥Ñ Raceleader ©ó ¬P´Á¤é ¥|¤ë 06, 2003 7:29 pm


§Aªº¥u­­¤E¦ì¼Æ¦r,¦Ó§Úªº¬O¤@¯ë¦¡
¨º§A»{¬°½ÖªºÃÒ©ú¦³»¡ªA¤O  °Ý¸¹

Raceleader
³X«È
 

¥Ñ kevin ©ó ¬P´Á¤é ¥|¤ë 06, 2003 8:47 pm


I think Raceleader is right...........

kevin
±Ð¡@±Â
±Ð¡@±Â
 
¤å³¹: 1158
µù¥U®É¶¡: 2002-12-22

¥Ñ scsnake ©ó ¬P´Á¤é ¥|¤ë 06, 2003 8:59 pm


¥i¤£­n°w¾W¬Û¹ï³é¡ã¡ã

scsnake
³X«È
 

¥Ñ Raceleader ©ó ¬P´Á¤é ¥|¤ë 06, 2003 9:08 pm


I just back from dinner :P

Raceleader
³X«È
 

¥Ñ Tassader-VIII ©ó ¬P´Á¤@ ¥|¤ë 07, 2003 7:15 pm


Raceleader ¼g¨ì:§Aªº¥u­­¤E¦ì¼Æ¦r,¦Ó§Úªº¬O¤@¯ë¦¡
¨º§A»{¬°½ÖªºÃÒ©ú¦³»¡ªA¤O  °Ý¸¹

§Ú»{¿é........ ®`²Û

Tassader-VIII
³X«È
 

¥Ñ ¯Á¬¥¦è ©ó ¬P´Á¤@ ¥|¤ë 07, 2003 7:19 pm


Tassader-VIII ¼g¨ì:
Raceleader ¼g¨ì:§Aªº¥u­­¤E¦ì¼Æ¦r,¦Ó§Úªº¬O¤@¯ë¦¡
¨º§A»{¬°½ÖªºÃÒ©ú¦³»¡ªA¤O  °Ý¸¹

§Ú»{¿é........ ®`²Û

You  lose !!!
”FÃC¤@®i¥u¬°§g
©]¤ë¦p©p¾K§Ú¤ß
¥»·í¤ß§NµL¤@ª«
©R©w¬°©p°Ê¯u±¡

¯Á¬¥¦è
ª©¡@¥D
ª©¡@¥D
 
¤å³¹: 2266
µù¥U®É¶¡: 2003-03-16
¨Ó¦Û: ¬õÅ]À]®É­p¥x

¥Ñ Raceleader ©ó ¬P´Á¤@ ¥|¤ë 07, 2003 8:32 pm


Don't be sad, we are learning here  £~£~£~

Raceleader
³X«È
 




¥N¼Æ¾Ç