又在打最後一句時按錯了制.....懊惱極了....#ed_op#br#ed_cl##ed_op#br#ed_cl#
txreformer 寫到:#ed_op#div#ed_cl##ed_op#strong#ed_cl#1.#ed_op#/strong#ed_cl# 582除以一個數,所得之不完全商為11,並且除數與餘數的差為6,請問除數、餘數各是多少?#ed_op#/div#ed_cl##ed_op#div#ed_cl#
#ed_op#br#ed_cl#設餘數為x, 除數即為x+6 [註:除數必大於餘數]#ed_op#br#ed_cl#582 = 11(x+6) + x = 12x + 66 > x = 43; x+6 = 49#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl#
txreformer 寫到:#ed_op#br#ed_cl##ed_op#strong#ed_cl##ed_op#/strong#ed_cl##ed_op#strong#ed_cl#2.#ed_op#/strong#ed_cl# 有一個數,除以7餘2,除以8餘4,除以9餘3,這個數是多少?#ed_op#/div#ed_cl##ed_op#div#ed_cl#
#ed_op#br#ed_cl#設 8*9*i = 1 (mod 7); 7*9*j = 1 (mod 8); 7*8*k = 1 (mod 9) => i=4, j=7, k=5;#ed_op#br#ed_cl#8*9*i *2 + 7*9*j *4 + 7*8*k *3 = 576+1764+840 = 156 (mod 7*8*9)#ed_op#br#ed_cl#此數為 504n+156, n∈#ed_op#span style="font-weight: bold; font-style: italic;"#ed_cl#N#ed_op#/span#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#/div#ed_cl#
txreformer 寫到:#ed_op#br#ed_cl##ed_op#strong#ed_cl##ed_op#/strong#ed_cl##ed_op#strong#ed_cl##ed_op#/strong#ed_cl##ed_op#div#ed_cl##ed_op#strong#ed_cl#3.#ed_op#/strong#ed_cl# 63285和70352的積除以7,餘數是多少?#ed_op#/div#ed_cl##ed_op#div#ed_cl#
#ed_op#br#ed_cl#63285 = 5 (mod 7); 70352 = 2 (mod 7); [註:63,28,7,35皆為7之倍數,數此二數能速算出]#ed_op#br#ed_cl#5*2 = 10 = 3 (mod 7) => 餘數為3#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#/div#ed_cl#
txreformer 寫到:#ed_op#br#ed_cl##ed_op#strong#ed_cl##ed_op#/strong#ed_cl##ed_op#strong#ed_cl##ed_op#/strong#ed_cl##ed_op#div#ed_cl##ed_op#strong#ed_cl#4.#ed_op#/strong#ed_cl# 陽曆1992年1月1日是星期三,陽曆2004年1月1日是星期幾?#ed_op#/div#ed_cl##ed_op#div#ed_cl#
#ed_op#br#ed_cl#2004年-1992年 = 12年; 12年有12/4=3個閏年; 1年有 365日=52星期+1日;#ed_op#br#ed_cl#12*1 + 3 = 15 = 1 (mod 7); 故 2004年1月1日為星期三+1=星期四#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#/div#ed_cl#
txreformer 寫到:#ed_op#br#ed_cl##ed_op#strong#ed_cl##ed_op#/strong#ed_cl##ed_op#strong#ed_cl##ed_op#/strong#ed_cl##ed_op#div#ed_cl##ed_op#strong#ed_cl#5.#ed_op#/strong#ed_cl# 甲乙丙丁玩報數,甲報1;乙報2;丙報3;丁報4;丙報5;乙報6;甲報7;乙報8... ...,報2003這數的為誰?#ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#br#ed_cl# 本席算是乙 ,不知對不對#ed_op#br#ed_cl#
#ed_op#br#ed_cl#2003 = 5 (mod 2*(4-1)) => 5為丙報#ed_op#br#ed_cl#故乙為不對...#ed_op#br#ed_cl##ed_op#/div#ed_cl#