设六进制形式ABC,则有
36A+6B+C = 81C+9B+A
即
35A= 80C+3B
显然B能被5整除,且0≤B≤5,因此B = 5
35A = 80C + 15
7A = 16C + 3
A = (16C + 3)/7 = [14C + 2C+3)]/7 = 2C + (2C+3)/7
显然2C+3能被7整除,C=2、9……,又0≤C≤5,因此C = 2
则A = 5
六进制552,九进制255
十进制 = 5*36 + 5*6 + 2 = 212