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F(1852083) is over 387 000 digits long. Too big for a post.

]]>Now I've got a good proof, and I wouldn't be troubling MathsIsFun and ryos anymore You two, along with Mathsyperson have helped me a lot.

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Ryos' number ends in 86977990646384748721787109376

]]>(A power of 6 can never end in 8!)

Anyway, thanks a lot. You have been of great help. ]]>

Well, I've hit the limit. I typed 2^195312500 into bc, and it merrily spit "Runtime error (func=(main), adr=15): exponent too large in raise" back at me.

Thats okay, ryos.

So the bc cannot handle 58,794,922 digits!

I hope I am not too demanding.

When you have the time, can you run 6^1953125 and post the last 10 digits. That would be only 1,519,827 dcigits

Thanks, in advance.

Maybe tonight...

]]>I'll run 2^195312500 tonight...

... you hope!

]]>I let it run all night, so I don't know how long it took. I'll run 2^195312500 tonight...

]]>Your help is appreciated. You have gone one step further and given the last few digits of 2^7812500. I am grateful to you.

Logarithms can be of great help in calculating higher powers. You don't get the exact value, but a very close answer. However, the number of digits is not affected unless the power is very high.

mathsyperson wrote:

I think I said something of a similar nature a while ago. I can't find it though.

Is it here?

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