Its a test for a license in networking and I have to do some sort of programming with binary coding etc. I will need to work with numbers high numbers but will not be permitted a calculator and will be rigorously timed. If I can find a quick way to do 2^n then I can solve any problem in a minute or less (which usually takes me around 5 mins). The methods listed earlier is pretty much what I use to do, however, figuring 256 x 256 x .... takes too long. I actually had to memorize a few of them like 2^32 is 4,294,967 296 and that worked out fine but prone for mistake for stressful test day.
if the test is multiple choice, i would try to look for tricks that might let you decide the total number of digits in the product, what the last digit is, 2nd to last, etc.
you know that to do this on paper, you add the sum of 2*256, 5*256, but each product is placed 1 spot to the left,etc.
1)total # of digits: if you start with 256*256, the 1st of the 5 digit product will consist of 2*256->5, but also of 5*256->1, so the 1st digit will be 6. you may try to work out the 2nd digit as well: it will consist of 2*256->1, 5*256->2(5*2=0, add 2 from the 5*5), and 6*256->1, so the 2nd digit is 1+2+1=4. now from 64*** x 64*** you may get that 6x64 contributes 4 to the 1st digit, and the other digits do not contribute because 6>4. you may notice that when you started with 3 digits(256), the product had 5(or 2n-1) digits, because the first digit did not produce a 10. in the 2nd example when you start with 64***, the product would have 10(or 2n) digits, because 6*6>10
2)it is much easier to get the last digit of your products, but rarely useful. 256^n will always have 6 as it's last digit, just because 6*6 produces 6 as the last digit. but if you started with some number that ended in 4, for example 254*254, then the product ends in 6, and if you keep squaring it, you will again be stuck with 6.
