28 Apr '10 23:30

What's the units digit of 13^(13^2010)?

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Halfway29 Apr '10 14:021 edit

Why not 13^(13^(2010 mod 4) mod 4) = 13^(13^2 mod 4) = 13.*Originally posted by mtthw***Yes - it comes down to calculating 13^2010 mod 4. I'm sure there's some theorem or other that can help bring that down to something more manageable.**

That means 3 is the last digit...right? Maybe I'm missing something.- Joined
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podunk, PA29 Apr '10 14:06

I dont know anything about modular arithmatic, so looks like im done. Just out of curiosity I know 13^2010 units digit ends in 9, will that help?*Originally posted by mtthw***Yes - it comes down to calculating 13^2010 mod 4. I'm sure there's some theorem or other that can help bring that down to something more manageable.**- Joined
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converging to it01 May '10 12:365 edits

Hmm...I think I was wrong, it was late and having malgebra troubles...anyway, lemme try again*Originally posted by clandarkfire***I honestly have no idea how to do this...could you explain your thinking a little?**

Since log_{10}(a^(b^c))=(b^c)*log_{10}(a) then

c*log_{10}(b)+log_{10}(log_{10}(a))=log_{10}(log_{10}(a^(b^c)))

Therefore

log_{10}(log_{10}(13^(13^2010)))=2010*log_{10}(13)+log_{10}(log_{10}(13))~=2239

giving that the correct answer should, I think have been roughly

10^2239 digits- Joined
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Auschwitz01 May '10 16:491 editPretty much what I got...

Use base-10 logarithms ("log" rather than "ln".)

All two-digit integers have logs larger than or equal to 1, and less than 2.

All three-digit integers have logs larger than or equal to 2, and less than 3.

In general: all n-digit integers have logs larger than or equal to n-1, and less than n.

Let x = 13^2010

Then log x = log(13^2010)

By log laws, the right-hand side becomes 2010*log(13), or roughly 2239.026138...

So x has 2240 digits.

You can write x as 10^2239.026138...

Breaking the exponent into whole and fractional parts:

x = 10^2239 * [10^0.026138136742...]

which can be rearranged to

= [1.062033...] * 10^2239

in scientific notation.

Now suppose y = 13^x

Then log y = log(13^x)

By log laws, log y = x*log(13)

So log y = 1.18 * 10^2239

1.18 * 10^2239 digits...Holy Christ.

By comparison, there are roughly 8.0 * 10^80 atoms in the observable universe, (which is a spherical shape and has a radius of 46.5 billion light years or 2.73931666 × 10^23 miles.)- Joined
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converging to it01 May '10 17:03

Heh...out of laziness and the convention followed by my lecturers and maths software I always take 'log' to mean the natural log (inspite of 'ln'ðŸ˜‰ so I specify base 10 logs by log_{10}*Originally posted by clandarkfire***Pretty much what I got...**

Use base-10 logarithms ("log" rather than "ln".)

All two-digit integers have logs larger than or equal to 1, and less than 2.

All three-digit integers have logs larger than or equal to 2, and less than 3.

In general: all n-digit integers have logs larger than or equal to n-1, and less than n.

Let x = 13^2010

Then lo ...[text shortened]... a spherical shape and has a radius of 46.5 billion light years or 2.73931666 × 10^23 miles.)

But yeah, it's a bloody huge number! ðŸ™‚