Jump to content
BrainDen.com - Brain Teasers
  • 0

dividing infinity by 17


BMAD
 Share

Question

2 answers to this question

Recommended Posts

  • 0

I was going to ask which direction the exponentials are evaluated.

But obviously 3 is first raised to the 4th power; that result is then raised to the 5th power; and so on.

Only finite power towers can be evaluated right to left.

Unless you are using ^ to denote multiplication. In which the remainder is zero. ;)

Link to comment
Share on other sites

  • 0

Assuming exponentiation.

 

First: 3^4^(2k+1) = 13 (mod 17) for any positive integer k.

Fermat's little theorem states that for p = 17 (prime) a^17 = a (mod 17), which for a = 3 with gcd(3,17)=1 means 3^16=1 (mod 17).

So 3^4^(2k+1) = (3^4^2)^k * 3^4 = 1*3^4 = 13 (mod 17). 

I don't really like infinities but for any member in the series x_n = 3^4^5^.. ^n, (with n>=5) there exists a positive integer m such that 5^.. ^n = 2*m+1 (as it is odd). Therefore x_n = 13 (mod 17) using k=m in the lemma above.

So, the answer to the question would be 13 i.e. the series y_n = x_n (mod 17) converges to 13 as n goes to infinity.

 

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...