1.證明 53^103+103^53 能被39整除。
Rewrite the question as
53^103 + 103^53 = a (mod 39)
=>
{53^103 + 103^53 = a (mod 3)
{53^103 + 103^53 = a (mod 13)
=>
{ (-1)^103 + (1)^53 = a (mod 3)
{ (1)^103 + (-1)^53 = a (mod 13)
=> a=0
therefore, 53^103+103^53 is divisible by 39
3.若792能整除整數 13xy45z ，求數位x、y及z。
Rewrite the question as
13xy45z = 0 (mod 792)
=>
{13xy45z = 0 (mod 2)
{13xy45z = 0 (mod 3)
{13xy45z = 0 (mod 11)
=>
{z is an even number
{1+3+x+y+4+5+z = 3a , where a is an integer
{z-5+4-y+x-3+1 = 0
=>
{z is an even number ...(1)
{ x+y=-13-z+3a ...(2)
{ x-y = 3-z ...(3)
By putting a=0,1,2,3,... and z=0,2,4,6,8
Solving (1) (2) & (3)
We have
(x,y,z) = (4,1,0) (2,1,2) (0,1,4) (7,4,0) (5,4,2) (3,4,4) (1,4,6) (8,7,2) (6,7,4) (4,7,6) (2,7,8)
5.利用模 9 (modulo 9)或模 11 (modulo 11)協助，求以下算式中的未知的數位：
(a)51840 × 273581 = 1418243y040
(b)512 × 1y53125 = 1000000000
改了用 y ...易看一些
(a)
273581 = 0 (mod 11)
=> 51840 × 273581 = 0 (mod 11)
=> 1418243y040 = 11a (mod 11) , where a is an integer
=> 0-4+0-y+3-4+2-8+1-4+1 = 11a
=> -13 - y = 11a
=> y = 9 (put a = -2)
(b)
1000000000 = 10 (mod 11)
512 × 1y53125 = 6(7-y) (mod 11)
=> 6(7-y) = 10+11a , where a is an integer
=> y = 9 (put a = -2)
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To be continued...
[ 本帖最後由 kazuhiko 於 7/8/2007 06:25 PM 編輯 ] |
發表於 7/8/2007 05:53 PM 
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