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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,895

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.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**im really bored****Member**- Registered: 2005-05-12
- Posts: 76

89376 and 09376???

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**ganesh****Moderator**- Registered: 2005-06-28
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Are you sure?

I am happy if they are 89376 and 09376.

Please also tell me how you got those numbers.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,608

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

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**ganesh****Moderator**- Registered: 2005-06-28
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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.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

Wow, you discovered a new number pattern thingy!

School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?

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**im really bored****Member**- Registered: 2005-05-12
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ganesh I just used a calculater to figure it out so it is accurate

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**ganesh****Moderator**- Registered: 2005-06-28
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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.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,608

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.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,895

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.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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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.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
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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.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,895

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 !!!

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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Interesting property

Yes, 1787109376² = 3193759921787109376

Is there a class of numbers like this?

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**im really bored****Member**- Registered: 2005-05-12
- Posts: 76

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

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

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**Mr T****Member**- Registered: 2005-03-30
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i used excel 4 my maths, ICT and physics coursework.

I come back stronger than a powered-up Pac-Man

I bought a large popcorn @ the cinema the other day, it was pretty big...some might even say it was "large

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**mathsyperson****Moderator**- Registered: 2005-06-22
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Happy 1000th Post!!!

Why did the vector cross the road?

It wanted to be normal.

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**mathsyperson****Moderator**- Registered: 2005-06-22
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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.

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**mathsyperson****Moderator**- Registered: 2005-06-22
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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 non-negative integers)

*Last edited by mathsyperson (2005-07-04 05:10:11)*

Why did the vector cross the road?

It wanted to be normal.

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**Roraborealis****Member**- Registered: 2005-03-17
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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?

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**mathsyperson****Moderator**- Registered: 2005-06-22
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I think orange probably set that record already at some point.

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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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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**ganesh****Moderator**- Registered: 2005-06-28
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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.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**im really bored****Member**- Registered: 2005-05-12
- Posts: 76

25927

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