若A=(2+1)(2^2+1)(2^4+1)(2^8+1)…(2^64+1),则A-1996的末位数字是多少?

Question

若A=(2+1)(2^2+1)(2^4+1)(2^8+1)…(2^64+1),则A-1996的末位数字是多少?

Answer 1

A=(2-1)(2+1)(2^2+1)(2^4+1)…(2^64+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)…(2^64+1)
=(2^4-1)(2^4+1)(2^8+1)…(2^64+1)
=(2^8-1)(2^8+1)…(2^64+1)
=(2^16-1)(2^16+1)(2^32+1)(2^64+1)
=(2^32-1)(2^32+1)(2^64+1)
=(2^64-1)*(2^64+1)
=2^128-1
2^1=2
2^2=4
2^3=8
2^4=16
...
2^4n/10余6
所以2^128-1/10余5
15-6=9
所以原式的末尾数字为9.