
 ganesh
 Moderator
I need your help
Can anyone tell me what (1) the last 5 digits of 2^500 and (2) the last 5 digits of 2^2500 are? This is an important link in my work on numbers.
Character is who you are when no one is looking.
 ganesh
 Moderator
Re: I need your help
Are you sure? I am happy if they are 89376 and 09376. Please also tell me how you got those numbers.
Character is who you are when no one is looking.
 MathsIsFun
 Administrator
Re: I need your help
Puzzle or serious, ganesh?
It sounds tempting ....
(Have you been on holidays, im really bored?)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
 ganesh
 Moderator
Re: I need your help
I am pretty serious, Administrator. I discovered that 2^20 ends in 76. Thereafter, any number of the form 2^20n ends in 76. Similarly, any number of the form 2^100n ends in 376. Next, I had to know the last few digits of 2^500 and 2^2500. When I learnt that they are 9376 and 09376, I was excited. Because, any number of the form 2^500n would then have to end in 9376 and every 2^2500n would have to end in 09376. But I wanted to be sure about it, thats the reason I sought to know how it was obtained.
Character is who you are when no one is looking.
 Roraborealis
 Super Member
Re: I need your help
Wow, you discovered a new number pattern thingy!
School is practice for the future. Practice makes perfect. But  nobody's perfect, so why practice?
Re: I need your help
ganesh I just used a calculater to figure it out so it is accurate
 ganesh
 Moderator
Re: I need your help
The calculator on my PC (in the scientific mode) gives the complete output of a function of resultant about 40 digits. I wonder how you did that with a calculator. Tell us how.
Character is who you are when no one is looking.
 MathsIsFun
 Administrator
Re: I need your help
Some calculator! 150+ digit accuracy.
I know there are fullaccuracy calculation programs around, coz I had to design one in first year (all it could do was multiply). I recall I stored the digits in an array and just followed standrd multiply and carry rules or some such.
On the other hand, because it is 2, then the binary is: 100000000000000000... (etc, 500, or 2500 zeros)
Ganesh, if you want, I can try to find some javascript or something that does fullaccuracy computations.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
 ganesh
 Moderator
Re: I need your help
Thank you, Gentleman! Yes, I just want to generate the output of 2^500, 2^2500 and 2^12500. If possible, even 2^62500 and 2^312500 The last number of these would be about 100,000 digits.
Character is who you are when no one is looking.
 MathsIsFun
 Administrator
Re: I need your help
Aha!
2^500 = 3,273,390,607,896,141,870,013,189,696,827,599,152,216,642,046,043,064,789,483 ,291,368,096,133,796,404,674,554,883,270,092,325,904,157,150,886,684,127,560, 071,009,217,256,545,885,393,053,328,527,589,376
But to go much higher gets real timeconsuming.
But (again), if you are not interested in the higher order digits, then that may speed up the process.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
 ganesh
 Moderator
Re: I need your help
I really don't need to bother what the higher dgitis are. The last 10 digits would do. Thanks again, for the trouble taken and the time spent.
Character is who you are when no one is looking.
 ganesh
 Moderator
Re: I need your help
At some point, 2 raised to the power of 4 x 5^n would end 1787109376. Thereafter, every higher power of that number would end in 1787109376. This is because any power of a number having its last 10 digits as 1787109376 ends in 1787109376 !!!
Character is who you are when no one is looking.
 MathsIsFun
 Administrator
Re: I need your help
Interesting property
Yes, 1787109376² = 3193759921787109376
Is there a class of numbers like this?
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
Re: I need your help
I just used a program, like what the administrator was talking about, that I made myself
Re: I need your help
I tried Excel, but it went crazy on me for no reason. I'm trying to fix it now.
Why did the vector cross the road? It wanted to be normal.
 Mr T
 Super Member
Re: I need your help
i used excel 4 my maths, ICT and physics coursework.
I come back stronger than a poweredup PacMan I bought a large popcorn @ the cinema the other day, it was pretty big...some might even say it was "large Fatboy Slim is a Legend
Re: I need your help
Why did the vector cross the road? It wanted to be normal.
Re: I need your help
Roraborealis wrote:Wow, you discovered a new number pattern thingy!
He missed out the first one: any number of the form 2^4n ends in 6.
Why did the vector cross the road? It wanted to be normal.
Re: I need your help
Sorry for the triple post, but while I was trying to fix my excel thingy, and failing, I discovered a generalisation of ganesh's discovery.
Any number of the form 2^((4*5^x)n+c) always has the same last (x+1) digits.(x and c are both nonnegative integers)
Last edited by mathsyperson (20050705 03:10:11)
Why did the vector cross the road? It wanted to be normal.
 Roraborealis
 Super Member
Re: I need your help
I think that's a new record. Triple post.
School is practice for the future. Practice makes perfect. But  nobody's perfect, so why practice?
Re: I need your help
I think orange probably set that record already at some point.
Why did the vector cross the road? It wanted to be normal.
 MathsIsFun
 Administrator
Re: I need your help
mathsyperson wrote:Any number of the form 2^((4*5^x)n+c) always has the same last (x+1) digits.(x and c are both nonnegative integers)
I don't have time right now to digest that, but it looks cool.
But I got to thinking that multiplying by 2 has the obvious pattern in the final digit: 2,4,8,6,2,... , so once it gets into a pattern, it can't stop. And that idea would apply to larger and larger series of final digits.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
 ganesh
 Moderator
Re: I need your help
Higher powers of numbers ending in 0,1,5, and 6 end in themselves. Of these numbers, I was most interested in 6 because numbers of the form 2^4n end in 6. For that matter, the last digits of higher powers rotate in cycles of 4, thus, they can easily be known. For example, try to figure out what the last digit of 23^611 would be. I shall help you with that if there are no answers.
Character is who you are when no one is looking.
