The answer is written everywhere

Algebra Level 3

$\Large 2^{2^{2^{2^2}}}$

If I were to insert a pair of parentheses into the number above, the resultant value might differ. For example, by adding the parentheses like

$\Large 2^{2^{2^{\left(2^2\right)}}},$

we would have the same result. But if I were to add the parentheses like

$\Large \left(2^2\right)^{2^{2^2}},$

the resultant number would not remain the same.

If we're only allowed to add exactly one pair of parentheses as shown above, find the minimum value (call it $m$ ) and the maximum value (call it $M$ ) of all the possible resultant numbers.

Then what is $\sqrt[m]{M}?$

The answer is 2.

1 solution

Hjalmar Orellana Soto
Mar 4, 2016

$M=2^{2^{2^{2^2}}}=2^{2^{16}}$ and $m=(2^{2^{2}})^{2^{2}}=2^{4\cdot 4}=2^{16}$ . Therefore: $\sqrt[m]{M}=2^{\frac{2^{16}}{2^{16}}}=2$

Well technically, you need to use ONE pair of parentheses for $M$ as well.

Thank you.

Pi Han Goh - 5 years, 3 months ago

The title gave away the answer please change it

Department 8 - 5 years, 3 months ago

It's intentional. I'm not changing it.

Pi Han Goh - 5 years, 3 months ago

@Pi Han Goh – Yes, I agree with Lakshya. The title makes it too obvious. :P

Mehul Arora - 5 years, 3 months ago

@Pi Han Goh – The answer is written 15 times!

Joel Yip - 5 years, 3 months ago

I think m should be 2³². Please check it. I'm not convinced at all.

Prayas Rautray - 3 years, 11 months ago

In the second step why should you multiply 2² with 2². 2² should be raised to the power 2². Check it again.

Prayas Rautray - 3 years, 11 months ago

I got it. Sorry for all the fuss!!!

Prayas Rautray - 3 years, 11 months ago

The answer is written everywhere

The answer is 2.

1 solution

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