Question

试确定(2+1)(2²+1)(2的4次方+1)(2的8次方+1)(2的16次方+1)(2的32次方+1)+1的

Answer 1

设S=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1

S=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1

=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1

=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1

=(2^8-1)(2^8+1)(2^16+1)(2^32+1)+1

=(2^16-1)(2^16+1)(2^32+1)+1

=(2^32-1)(2^32+1)+1

=2^64-1+1

=2^64

=(2^4)^16

=(16)^16

S个位数是6 （16个16相乘个位数总是6）