When the expression \((2^{1})(2^{2})(2^{3})...(2^{99})(2^{100})\) is written as an integer, what is the product of the tens digit and the ones digit?
Note that (21) (22) (23) (24)......(299)(2100) can be rearranged as
(21)(2100) * ( 22)(299) * ( 23)(298) =
(2101)50 =
(250)101 =
( [(2^10)]^5)^101
Note that 2^10 ends in 24
And 24^(2n + 1) ends in 24 for n = any whole number......so we can write
[( ......24)^5]^101 =
[ (......24)^(2(2) + 1) ]^101 =
[(.....24) ] ^(2(50) + 1) =
24
And the product of the tens digit and ones digit = 8
These exponents (2^1) (2^2) (2^3) (2^4)......(2^99)(2^100) can be written thus:
2^(100*101/2) = 2^5050. This, in turn, can be split into: 2^5045 x 2^5. And since 2 to any exponent ending in 5 will result ending in 32. Therefore, 2^5045 x 2^5 =.......32 x 32 =1,024, or......24. And their product = 2 x 4 = 8.