At a first glance, it seems that I need to do this:


But afterwards, I can't find an algebric manipulation that will lead me to the solution, I took the exercise out of one of the entry tests of TAU.

  • 2
    $\begingroup$ You can of course just calculate it by multiplying out the brackets. The result is 64. I guess you want a different way of doing it. Any requirements? $\endgroup$ – Andreas Sep 8 '18 at 17:12
  • 1
    $\begingroup$ This: $$\prod_{k=0}^{n}{\bigg[2^{2^k}+3^{2^k}}\bigg]<3^{2^{n+1}}$$ might help. $\endgroup$ – Rhys Hughes Sep 8 '18 at 17:17
  • $\begingroup$ There is a neat way of doing this. It would work with other numbers in a similar set of relationships and doesn't in general depend on the fact that $3-2=1$ $\endgroup$ – Mark Bennet Sep 8 '18 at 17:20

Since \begin{align*} &\phantom{==}(3-2)(3+2)(3^2+ 2^2)(3^4+2^4)\cdots(3^{32}+2^{32})\\ &= (3^2-2^2)(3^2 + 2^2)(3^4+2^4)\cdots(3^{32}+2^{32})\\ &= (3^4-2^4)(3^4+2^4)(3^8+2^8)\cdots(3^{32}+2^{32})\\ &= \cdots\\ &= 3^{64}-2^{64}, \end{align*} the result is $\log_3(3^{64}) = 64$. The trick is $(a-b)(a+b)= a^2 - b^2$.

  • $\begingroup$ Awesome, I am feeling pretty stupid not able to see this. $+1$ $\endgroup$ – paulplusx Sep 8 '18 at 17:27
  • 3
    $\begingroup$ The app and the site always have the delay. I have encountered several times. $\endgroup$ – xbh Sep 8 '18 at 17:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.