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Is a large integer divisible by 2^n? 2017-03-28
From Sahand:
You are given number x that is very large (that large that it can't be divided by hand) can we find out that x is divisible by 2^n or not?
Answered by Penny Nom.
a^(2^n) 2010-10-22
From Tim:
I am trying to understand a^(2^n). The hint they give is a^(2^(n+1)) = (a^(2^n))^2 I am writing a program that will solve a^(2^n) recursively but need to understand the power before I begin. I am currently pursuing writing (a) x (a^(2^(n-1))) where the (a^(2^(n-1))) would be the recursive function call a n approaches 0. Once n is 0, the result would be multiplied by a two more times. Anyway, explaining these powers would be appreciated. I will most likely complete the program before the answer but I want to understand the logic of these powers. Thank you, Tim
Answered by Stephen La Rocque.
The number of decimal places in 1 over a power of 2 2002-09-12
From Allan:
Does anyone notice that the maximum number of decimal place of the number 2 dividing 1 and its increment (4, 8, 16...etc) is the same as the power of number 2? eg. 22=4, thus the max number of decimal of 1/4=0.25 which is 2 decimal place and 2 is the number of power of 2 take 64 as example: 26=64, and take 1/64=0.015625 which has 6 decimal place (and is the power 6)

Is there such a law in math? If yes, can you tell me what it is? Or is this my discovery?

Answered by Paul Betts.



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