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$$\log_3(5(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})+2^{64})=?$$

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

$$\log_3((2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})+2^{64})=?$$

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.

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    $\begingroup$ 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, 2018 at 17:12
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    $\begingroup$ This: $$\prod_{k=0}^{n}{\bigg[2^{2^k}+3^{2^k}}\bigg]<3^{2^{n+1}}$$ might help. $\endgroup$
    – Rhys Hughes
    Sep 8, 2018 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, 2018 at 17:20

1 Answer 1

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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 $(a-b)(a+b)= a^2 - b^2$.

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  • $\begingroup$ Awesome, I am feeling pretty stupid not able to see this. $+1$ $\endgroup$
    – paulplusx
    Sep 8, 2018 at 17:27
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    $\begingroup$ The app and the site always have the delay. I have encountered several times. $\endgroup$
    – xbh
    Sep 8, 2018 at 17:37

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