整数的同余与尾数特征问题一个大于2002的奇数恰能被33整除,这个数加上1,就能被4整除,加上2,就能被5整除,减去3,就能被6整除,则满足条件的最小数是?

Question

整数的同余与尾数特征问题
一个大于2002的奇数恰能被33整除,这个数加上1,就能被4整除,加上2,就能被5整除,减去3,就能被6整除,则满足条件的最小数是?

Answer 1

x=33k
x>2002
奇数
k=2p+1
x=33(2p+1)=66p+33
x+1Ξ0(mod4)
x+2Ξ0(mod5)
x-3Ξ0(mod6)
x-3Ξ0(mod4)
x-3Ξ0(mod5)
x-3Ξ0(mod6)
x-3Ξ0(mod[6,5,4])Ξ0(mod60)
x=60m+3
S=60m+3=66p+33
60m+3=66p+33
5m+3Ξ0(mod11)
60m+3=66p+33
20m+1＝22p＋11
m=6
p=10
是特解
m=6+11s
p=10－10s
S=60m+3＝60(6+11s)+3
=363+660s
s=3
S=2343