Are you sure?
I am happy if they are 89376 and 09376.
Please also tell me how you got those numbers.

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.

ganesh I just used a calculater to figure it out so it is accurate

Some calculator! 150+ digit accuracy.

I know there are full-accuracy 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 full-accuracy computations.

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 time-consuming.

But (again), if you are not interested in the higher order digits, then that may speed up the process.

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 !!!

I just used a program, like what the administrator was talking about, that I made myself

i used excel 4 my maths, ICT and physics coursework.

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.

I think that's a new record. Triple post.

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 non-negative 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.

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