Phân tích mỗi số ra TSNT:
68 = 22 x 17
264 = 23 x 3 x 11
15 = 3 x 5
=> BCNN (68;264;15) = 23 x 5 x 3 x11 x17 = 22440
=> BCNN (68;264;15) = B(22440) = { 0 ; 22440 ; 44880 ; 67321 ; 89760 ; .......}
=> UCLN (68;264;15) = 1
=> UC (68;264;15) = 1
Ta có: 68 = 22 . 17
264 = 23. 3 . 11
15 = 3 . 5
=> BCNN(68, 264, 15) = 23 . 3 . 5 . 11 . 17 = 22440
=> ƯCLN(68, 264, 15) = 1
Khi đó:
BC (68, 264, 15) = B(22440) = {0 ; 22440 ; 44880 ; 67320 ; ...}
ƯC(68, 264, 15) = Ư(1) = {1}