Originally posted by mtthwWhy not 13^(13^(2010 mod 4) mod 4) = 13^(13^2 mod 4) = 13.
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.
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Originally posted by mtthwI 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?
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 clandarkfireHmm...I think I was wrong, it was late and having malgebra troubles...anyway, lemme try again
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
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 clandarkfireHeh...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}
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! 🙂