You know me!

Algebra Level 3

What is the value of this expression (up to 3 decimal places)?

$\large \left(1+9^{-4^{7 \times 6}}\right)^{3^{2^{85}}}$

This problem is not original.

The answer is 2.718.

1 solution

Vinayak Srivastava
May 4, 2020

The given expression is :

${{{(1+9^{{-4}^{7×6}})}^3}^2}^{85}$

$9^{{4}^{7×6}}=9^{{4}^{42}}={{3}^2}^{85}$

$\displaystyle \lim_{n \to \infty}\left({{1+\dfrac{1}{n}}}\right)^n = e$

Hence,

${{{(1+9^{{-4}^{7×6}})}^3}^2}^{85}=2.718$ (upto 3 decimal places).

Video link

Your last line isn't true. $e$ is a transcendental number , so it can't be expressed algebraically.

Pi Han Goh - 1 year, 1 month ago

I have edited the last line. Thanks for your comment!

Vinayak Srivastava - 1 year, 1 month ago

I got to admit: This is a very cute question!

For completeness, it would be better to show that $f(x) = (1 + \frac1x)^x = 2.718$ to 3 decimal places for $x \geqslant 3^{2^{85}}$ (or smaller).

Do you know how to complete this?

Pi Han Goh - 1 year, 1 month ago

We know that :

$\lim_{n \to \infty}({{1+\frac{1}{n}}})^n = e$

This is a definition of the mathematical constant ' $e$ '.

As ' $n$ ' "approaches" infinity, the expression gets closer and closer to ' $e$ '.

However, if we want only some first digits of ' $e$ ', even a small number like ' $5000$ ' put in place of ' $n$ ' would give

$({{1+\frac{1}{5000}}})^{5000}=2.718$ (upto 3 decimal places).

In my opinion, the beauty of the expression ${{{(1+9^{{-4}^{7×6}})}^3}^2}^{85}$ is that it uses all the numbers $1$ to $9$ that too only once (Pandigital number) and it has been found surprisingly accurate since ${{3}^2}^{85}$ is a very large number.

Vinayak Srivastava - 1 year, 1 month ago

@Vinayak Srivastava – Yup, that's it! Wonderful!

Pi Han Goh - 1 year, 1 month ago

I believe that your second line is not accurate $(9^{4^{42}} \ne 3^{2^{85}})$ but instead $(9^{4^{42}} = 3^{\color{#3D99F6}{4}^{\color{#D61F06}{84}}})$

Mahdi Raza - 1 year, 1 month ago

${{3}^2}^{85}=9^{{2}^{84}}=9^{{4}^{42}}=9^{{4}^{7×6}}$

Vinayak Srivastava - 1 year, 1 month ago

$9^{4^{7*6}}=3^{2^{4^{42}}}$ . So $2^{168}=2^{85}$ ? $168=85$ ?

A Former Brilliant Member - 11 months, 3 weeks ago

@A Former Brilliant Member – When you take it to be $3^2$ , you need to put parenthesis.

${9^{{4}^{42}}} = {(3^2)^{{4}^{42}}}$ which is not equal to ${3^{2^{{4}^{42}}}}$

A common example to see this is:

$2^{3^{4}}=2^{81} \neq (2^3)^4=2^{12}$

Vinayak Srivastava - 11 months, 3 weeks ago

@Vinayak Srivastava – Ok. Thank you!

A Former Brilliant Member - 11 months, 3 weeks ago

@A Former Brilliant Member – You're welcome! This question even confused me a lot. It was copied from some video on Numberphile and this being my first post, I forgot to mention it. I'll add that. Thanks!

Vinayak Srivastava - 11 months, 3 weeks ago

You know me!

The answer is 2.718.

1 solution

0 pending reports