# Googol and Googolplex Question(Australian Maths Competition)

Not sure if I should ask this here, but I've tried other places with no success. I've pretty lost with this question..

Question: One googol is the number G=10¹⁰⁰, and one googolplex is 10ᴳ. Let n be the largest whole number for which nⁿ<10ᴳ. How many digits does n have?

• When was this competition? Most importantly: is it still ongoing? Commented Jul 31, 2017 at 7:15
• @Arthur Australian Maths Competition, 27th July. It is not ongoing I believe..That day only. Commented Jul 31, 2017 at 7:24
• Seems you're right (although I can't find the actual problems to confirm). Commented Jul 31, 2017 at 7:27

I think it can be solved in this way:

We have the condition $n^n<10^G$

Since we are talking about a large number, it is safe to do: $\log_{10}(n^n) < \log_{10} (10^G)$

Simplifying a little gives us $n\log_{10}(n) < 10^{100}$ since $G = 10^{100}$

Taking log again on both sides will give us: $\log_{10}(n\log_{10}(n)) < 100$ which can be written as $\log(n) + \log(\log(n)) < 100\ \ ...(1)$

Now let $x$ be the number of digits in the number $n$, then $x \approx \log_{10}(n)$

We can rewrite equation 1 above as $x + \log(x) < 100$ for which the largest value of x that satisfies the equation is 98.

Edit: (correction as suggested by Arthur)

The number then would be $n \approx 1.02 \times 10^{98}$ which will then have 99 digits. So the answer would be 99

• if $x = 99$, then $logx+x = 100.99$ which does not satisfy the equation right? Commented Jul 31, 2017 at 7:30
• oh yeah.. you are right.. $10^{98}$ itself would have 99 digits. I'll make the required changes. Thanks for pointing it out Commented Jul 31, 2017 at 7:35
• As a side note to @Jonathan, this answer points to the fact that when deciding how large a power expression is, the exponent is far more important than the base. In this case, for the number $10^G$, if you multiply the base by (about) $10^{97}$, you only need to divide the exponent by (about) $100$ to "make up" for it. Commented Jul 31, 2017 at 7:43