FANDOM


Okay, the Second Googological Olympiad has officially ended.  Four people sent in solutions, and two sent in solutions to all six problems.  The winner is Wojowu, who scored a stellar 47 out of 50 points. On April 15, a person named January First-of-May sent in solutions; while technically his scores should not count, I don't see a problem with adding his scores to the list.


Results of the Second Olympiad
NameP 1P 2P 3P 4P 5BonusTotal
Wojowu55101010747
Wythagoras55596131
Cloudy17655    10
January First-of-May55    10
cookiefonster5     5


Thanks to everyone who participated! I had a lot of fun writing and overseeing the contest.


Solution to Problem #1:

The answer is \(\prod_{i=0}^{n-1} 10[3]i\), where we take \(10[3]0\) to be \(10\). In fact \(10[3]n = 10^{\prod_{i=0}^{n-1} 10[3]i}\), which we can prove this by induction.

Base case: \(10[3]1 = 10^{10} = 10^{10[3]0}\).

Inductive step: Assume \(10[3]n = 10^{\prod_{i=0}^{n-1} 10[3]i}\). Then

\(10[3](n+1) = (10[3]n)^{10[3]n} = (10^{\prod_{i=0}^{n-1} 10[3]i})^{10[3]n} = 10^{(\prod_{i=0}^{n-1} 10[3]i) 10[3]n} = 10^{\prod_{i=0}^n 10[3]i}\).


Solution to Problem #2:

We use the following inequalities for the pth prime \(p_n\):

\(n (\log n + \log (\log n) - 1) < p_n < n (\log n + \log (\log n))\) for \(n \ge 6\).

So if we let \(N\) be the googolplexth prime,

\(N > 10^{10^{100}}(\log(10^{10^{100}}) + \log(\log(10^{10^{100}})) - 1) > 10^{10^{100}}(\log(10^{10^{100}}) + 100 - 1)\\ > 10^{10^{100}}10^{100}\log 10 > 10^{10^{100}}10^{100}(2.302585092994)\)

\(N < 10^{10^{100}}(\log(10^{10^{100}}) + \log(\log(10^{10^{100}}))) < 10^{10^{100}}(\log(10^{10^{100}}) + \log(10^{101}))\\ < 10^{10^{100}}(10^{100}\log(10) + 300) < 10^{10^{100}}10^{100}(2.302585092995 + .000000000001)\)

Since multiplying a number by \(10^{10^{100}}10^{100}\) does not change its digits, the first 10 digits of \(N\) are \(2302585092\).


Solution to Problem #3:

Lower bound:

Since the function \(x \log x\) is convex (one can check that the second derivative is \(\frac{1}{x}\), which is positive for \(x > 0\)), the integral from \(1\) to \(n\) will be less than its approximation using the trapezoidal rule, so we have

\(\int_1^n x \log x \, dx < \sum_{i=1}^{n-1} \frac{i \log i + (i+1)\log(i+1)}{2}\)

Therefore

\(\log H(n) = \sum_{i=1}^n i \log i > \sum_{i=1}^n i \log i - \sum_{i=1}^{n-1} \frac{i \log i + (i+1)\log(i+1)}{2} + \int_1^n x \log x \, dx\\ = \sum_{i=1}^n i \log i - (\sum_{i=1}^{n-1} i \log i + \frac{n \log n}{2}) + (\frac{x^2 \log x}{2} - \frac{x^2}{4})_{x=1}^n\\ = \frac{n \log n}{2} + \frac{n^2 \log n}{2} - \frac{n^2}{4} + \frac{1}{4}\)

\(H(n) > e^{\frac{n^2 \log n + n \log n}{2} - \frac{n^2-1}{4}} = \frac{n^{\frac{n^2+n}{2}}}{e^{\frac{n^2-1}{4}}}\).

Upper bound:

We can write

\(H(n) = (1 - \frac{1}{2})^1 (1 - \frac{1}{3})^3 \ldots (1 - \frac{1}{n})^{\frac{n(n-1)}{2}} n^{\frac{n(n+1)}{2}}\)

which we can verify by induction.

Base case: \(H(1) = 1\)

Inductive step: Assume \(H(n) = (\prod_{i=2}^n (1 - \frac{1}{i})^{\frac{i(i-1)}{2}}) n^{\frac{n(n+1)}{2}}\).

\(H(n+1) = H(n)(n+1)^{n+1} = (\prod_{i=2}^n (1 - \frac{1}{i})^{\frac{i(i-1)}{2}}) n^{\frac{n(n+1)}{2}} (n+1)^{n+1}\\ = (\prod_{i=2}^n (1 - \frac{1}{i})^{\frac{i(i-1)}{2}}) (1 - \frac{1}{n+1})^{\frac{n(n+1)}{2}} (n+1)^{\frac{n(n+1)}{2}} (n+1)^{n+1}\\ = (\prod_{i=2}^{n+1} (1 - \frac{1}{i})^{\frac{i(i-1)}{2}}) (n+1)^{\frac{(n+1)(n+2)}{2}}\).

We use the inequality \((1 - \frac{1}{i})^i < \frac{1}{e}\)*. Then

\(H(n) = (\prod_{i=2}^n (1 - \frac{1}{i})^{\frac{i(i-1)}{2}}) n^{\frac{n(n+1)}{2}} \\ < (\prod_{i=2}^n e^{-\frac{i-1}{2}})n^{\frac{n(n+1)}{2}} = e^{-\frac{n^2-n}{2}}n^{\frac{n(n+1)}{2}} = \frac{n^{\frac{n(n+1)}{2}}}{e^{\frac{n^2-n}{2}}}\)

*to verify the inequality \((1 - \frac{1}{i})^i < \frac{1}{e}\), observe that \(\lim_{i \to \infty} (1 - \frac{1}{i})^i = \frac{1}{e}\), and then observe that \((1 - \frac{1}{i})^i\) is increasing (the first derivative is positive).

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