Power of Power
posted by Rajesh Lal at
9:17 AM
Saturday, November 29, 2008
Links
Search
Share
- Tom Chrome Extension
- ToM @ Twitter
- MeeT ToM
- Daily Tricks Widget
- Vista Gadget 1.5
- iPhone Widget
- Windows XP Widget
- Google Toolbar Button
- Suggestions
Favorite Links
- Vista Sidebar Gadgets
- Widgets for Web 2.0
- C Sharp Tricks
- Interview Information
- Illustrator - Elar Alexander
Bookmark Us
Previous Posts
- As easy as 1 2 3.....
- Cannon Ball Conundrum
- More Rebuses
- Placing Riddle
- Black Friday
- Self Contained
- Maths with Zero
- Breakfast Blues
- 3 barrels
- Consecutive double letters
14 Comments:
Method
Step 1: Post eguation on a popular puzzle blog.
Step 2: Wait...
Whatever is the anti-logarithm of
439225223.11307363290327076136079
its very simple, but unfortunatly i must be off outside to get some fresh air AND A LIFE
The anonomous above is on the right track.
You take the log of 2890978
and multiply it by 67980526
the answer is
=exp(ln(2890978)*67980526)
Or you use another base.. using base 10, the answer is
10^(log(2890978)*67980526)
But then this is difficult to evaluate also.
Now you could find factors or
67980526 like 33990263 *2, this
would help.
But I had no luck going further.
Could 33990263 be prime?
hjg
An approximation would be 1.297 * 10^439225223. You can get the mantissa by raising 10 to the power of the fractional part of the log base 10 of the answer, which can be calculated with
67980526 * ln 2890978 / ln 10
on a good calculator and was given in an earlier answer. The exponent, of course, is the integer part of the above.
But I assume you want the exact answer... Since it is a half a billion digits long or so, you will need:
- a very long piece of paper for the answer (about 19 miles, I estimate, if you get 50 lines of 80 characters per regular page)
- a multi-precision arithmetic package, such as GNU MP to represent really long numbers
- a fast multiplication algorithm for long numbers, GNU MP implements Schönhage-Strasse, O(N*logN) performance by utilizing the Fast Fourier Transform (FFT) to enable doing multiplication digit-by-digit without all the carries and then adjusting the result.
- a fast exponentiation algorithm based on multiplication. One is built in to GNU MP, but it is simple to put one together, if you notice that, for example,
a^13 = ((a^2*a)^2)^2*a,
so you can do exponentiation in O(N) by repeated squaring and multiplication.
GNU MP (via a perl module in my case) provides the answer after about a minute of crunching on my 2.4 GHz CPU. The last few digits of the answer are ...11835542421504.
If you would like the complete result, please let me know the e-mail address to send it to :)
To Surge
Wow!
I'll check that, but I think it may take some time.
hjg
I think that the 3rd anonymous was right GET A LIFE
by the way is this a homework problem of yours or somthing?
use a calculater
To Surge
I wanted to use some factors like you spoke of in
a^13 = ((a^2*a)^2)^2*a
but after 67980526 = 33990263 *2
I could not find any other factors
and I searched up to 2000.
I'll try again.
hjg
hjg,
the factorizations for the two numbers are:
2890978 = 2 * 71 * 20359
67980526 = 2 * 33990263
The big factors above can be checked for being prime in any number of ways, the easiest of which is to Google 'prime number checker', which will get you to a number of pages, that offer an online checker. There are some cool ways of doing this described in the article on prime numbers in Wikipedia, under primality tests. Miller-Rabin, in particular, is widely documented and implemented. AKS is very recent and appears to be stunningly superior to others.
Sorry for the diversion offtopic.
You went about half way in your proof, since if 33990263 had a factor, it would have to be less than int(sqrt(33990263)) = 5830
Yeah, I factored them first and thought to take advantage of that, but since they both have huge factors, it really did not go anywhere. I am suspicious of these large factors, as this seems hardly random and I suspect could be used somehow, but my number theory knowledge is too childish to go anywhere with that.
Maybe someone else will clue us in in this direction...
Type "is 33990263 prime?" in google. Google is amazing.
multiply the parintisised number by its self and use the other number as an exponet for te new number
who was the person who said they needed a life.its very simple u r depressed and obviously already think u never had a life. you dont take fun activities like these seriously cause u must be to stupid to work out a few f@#$%*g sums. shame i have pit 4 u as u probably battled to figure out that remark. boo hoo...stupid get a life, u seriously need 1 and fast. ;)
I ADMIRE THE LAST COMMENT EFT BY ANONYMOUS. ALL ANONYMOUS IS SAYING IS THAT THEY SUMS AND TRICK OF MIND IS TESTING US BECAUSE THEY KNOW THAT GOD PUT US ON THIS EARTH WITH A PURPOSE AND THAT WE ALL HAVE A LIFE AND A BRAIN. SO UTILISE IT AND CHERISH LIFE WHILE U ALIVE!! TEST YOURSELF WITH THESE SUMS... U CAN DO IT =)
Post a Comment
Links to this post:
Create a Link
<< Home