標題:
發問:
有無23既倍數係比118既倍數多1呀 更新: 即係23既倍數>118既倍數 1咁多
最佳解答:
圖片參考：http://hk.yimg.com/i/icon/16/35.gif 23x-1=118y x=2n-1 46n-23-1=118y 46n-24=118y 23n-12=59y 59y=(59,118,177,236,295,354,413,..............................) (59,118,177,236,295,354,413,..............................)+12 =71,130,189,248,307,366,425............. 70/23...........2 129/23..........15 188/23...........5+23=28 2,15,5,18,8,21,11,1,14,4,17,7,20,10,23 59*15=885+12=897 23n-12=59y 23x-1=118y n=39 y=15 x=2*39-1 x=77 23*77-1=1771-1 =1770
其他解答:
NO 118+1=119 119不能正整徐23 23*1=23 23*2=46 23*3=69 23*4=92 23*5=115 23*6=138 so 有眼看啦 沒有 或者是: 119/23=5+4/23 請大家投我一票|||||Since 8*118 - 23 * 41 = 1, 118 - 117 * (8*118 - 23 * 41 ) = 1 23 * (41 * 117) - (118 * 8) * 118 = 1 2006-11-17 23:37:15 補充： to find: 8*118 - 23 * 41 = 1118 - 23*5 = 323 - 3*7 = 2So, 23 - (118 - 23*5) *7 = 2 3 - 2 = 1[118 - 23*5] - [23 - (118 - 23*5) *7]=1118*8 - 23*41 = 1|||||1771=23 x 77 = 15 x 118 +1 2006-11-17 19:20:55 補充： the next one is 4485 = 23 x 195 = 118 x 38 1the next one is 7199 = 23 x 313 = 118 x 61 1the number are found by writing a very little program by myself 2006-11-17 19:22:25 補充： The general term is 23 x (77 118n) = 118 x (15 23n) 1for n is any positive integer|||||無23既倍數係比118既倍數多1|||||118÷23 = 115+1-2 =116 118-116|||||冇的... 最近的都只是23乘5.2=119.6
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maths送分題-------快!發問:
有無23既倍數係比118既倍數多1呀 更新: 即係23既倍數>118既倍數 1咁多
最佳解答:
圖片參考：http://hk.yimg.com/i/icon/16/35.gif 23x-1=118y x=2n-1 46n-23-1=118y 46n-24=118y 23n-12=59y 59y=(59,118,177,236,295,354,413,..............................) (59,118,177,236,295,354,413,..............................)+12 =71,130,189,248,307,366,425............. 70/23...........2 129/23..........15 188/23...........5+23=28 2,15,5,18,8,21,11,1,14,4,17,7,20,10,23 59*15=885+12=897 23n-12=59y 23x-1=118y n=39 y=15 x=2*39-1 x=77 23*77-1=1771-1 =1770
其他解答:
NO 118+1=119 119不能正整徐23 23*1=23 23*2=46 23*3=69 23*4=92 23*5=115 23*6=138 so 有眼看啦 沒有 或者是: 119/23=5+4/23 請大家投我一票|||||Since 8*118 - 23 * 41 = 1, 118 - 117 * (8*118 - 23 * 41 ) = 1 23 * (41 * 117) - (118 * 8) * 118 = 1 2006-11-17 23:37:15 補充： to find: 8*118 - 23 * 41 = 1118 - 23*5 = 323 - 3*7 = 2So, 23 - (118 - 23*5) *7 = 2 3 - 2 = 1[118 - 23*5] - [23 - (118 - 23*5) *7]=1118*8 - 23*41 = 1|||||1771=23 x 77 = 15 x 118 +1 2006-11-17 19:20:55 補充： the next one is 4485 = 23 x 195 = 118 x 38 1the next one is 7199 = 23 x 313 = 118 x 61 1the number are found by writing a very little program by myself 2006-11-17 19:22:25 補充： The general term is 23 x (77 118n) = 118 x (15 23n) 1for n is any positive integer|||||無23既倍數係比118既倍數多1|||||118÷23 = 115+1-2 =116 118-116|||||冇的... 最近的都只是23乘5.2=119.6
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