Math

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

Originally posted by clandarkfire
What's the units digit of 13^(13^2010)?

I can narrow it down to 3, or 7....having trouble getting it the rest of the way.

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.

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.

Why not 13^(13^(2010 mod 4) mod 4) = 13^(13^2 mod 4) = 13.

That means 3 is the last digit...right? Maybe I'm missing something.

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.

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?

The answer is three 🙂

Next question:

How many digits does 13^(13^2010) have? An aproximate is obviously fine; its a big number.

Originally posted by clandarkfire
The answer is three 🙂

Next question:

How many digits does 13^(13^2010) have? An aproximate is obviously fine; its a big number.

screwed up...I'll try again

Originally posted by Agerg
screwed up...I'll try again

I might be wrong but expressing 13 as 10^log_{10}(13 )I'm thinking its about 10^29107

I honestly have no idea how to do this...could you explain your thinking a little?

Originally posted by clandarkfire
I honestly have no idea how to do this...could you explain your thinking a little?

Hmm...I think I was wrong, it was late and having malgebra troubles...anyway, lemme try again

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

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 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.)

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.)

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'😉 so I specify base 10 logs by log_{10}

But yeah, it's a bloody huge number! 🙂