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.

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.

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. 2^{2}=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: 2^{6}=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.

Page 1/1

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.