The Angry Statistician

The Kelly Criterion is an alternative to standard utility theory, which seeks to maximize expected utility. Instead, the Kelly Criterion seeks to maximize expected growth . That is, if we start out with an initial bankroll $B_0$, we seek to maximize $\mathrm{E}[g(t)]$, where $B_t = B_0\cdot e^{g(t)}$. As a simple example, consider the following choice. We can have a sure $3000, or we can take the gamble of a $\frac{4}{5}$ chance of $4000 and a $\frac{1}{5}$ chance of $0. What does Kelly say? Assume we have a current bankroll of $B_0$. After the first choice we have $B_1 = B_0+3000$, which we can write as $$\mathrm{E}[g(1)] = \log\left(\frac{B_0+3000}{B_0}\right);$$ for the second choice we have $$\mathrm{E}[g(1)] = \frac{4}{5} \log\left(\frac{B_0+4000}{B_0}\right).$$ And so we want to compare $\log\left(\frac{B_0+3000}{B_0}\right)$ and $\frac{4}{5} \log\left(\frac{B_0+4000}{B_0}\right)$. Exponentiating, we're looking for the positive root of \[{\left({B_0+3000}\...

Find four prime divisors < 100 for $3^{32}-2^{32}$. Source: British Math Olympiad, 2006. This factors nicely as $3^{32}-2^{32} = \left(3^{16}+2^{16}\right)\left(3^{16}-2^{16}\right)$, and we can continue factoring in this way to get $$3^{32}-2^{32} = \left(3^{16}+2^{16}\right)\left(3^8+2^8\right)\left(3^4+2^4\right)\left(3^2+2^2\right)\left(3^2-2^2\right).$$ The final three terms are $5, 13, 97$, so we have three of the four required primes. For another prime divisor, consider $3^{16}-2^{16}$. By Fermat's Little Theorem $a^{16}-1\equiv 0 \bmod 17$ for all $a$ with $(a,17)=1$, and so it follows that $3^{16}-2^{16}\equiv 0 \bmod 17$, and we therefore have $17$ as a fourth such prime divisor. Alternatively, note $\left(\dfrac{3}{17}\right)=-1, \left(\dfrac{2}{17}\right)=1$, hence by Euler's Criterion $3^8\equiv -1 \bmod 17$ and $2^8\equiv 1 \bmod 17$, giving $3^8+2^8\equiv 0\bmod 17$.

Let $\left(1+\sqrt{2}\right)^{2012}=a+b\sqrt{2}$, where $a$ and $b$ are integers. What is the greatest common divisor of $b$ and $81$? Source: 2011-2012 SDML High School 2a, problem 15. Let $(1+\sqrt{2})^n = a_n + b_n \sqrt{2}$. I've thought about this some more, and there's a nice way to describe the highest power of $3$ that divides $b_n$. This is probably outside of the scope of the intended solution, however. First note that $(1-\sqrt{2})^n = a_n - b_n \sqrt{2}$, and so from $(1+\sqrt{2})(1-\sqrt{2})=-1$ we get $(1+\sqrt{2})^n (1-\sqrt{2})^n = {(-1)}^n$. This gives $${a_n}^2 - 2 {b_n}^2 = {(-1)}^n.$$ Now define the highest power of a prime $p$ that divides $n$ to be $\operatorname{\nu}_p(n)$. From cubing and using the above result it's straightforward to prove that if $\operatorname{\nu}_3(b_n) = k > 0$ then $\operatorname{\nu}_3(b_{3n}) = k+1$. Note $(1+\sqrt{2})^4 = 17 + 12\sqrt{2} \equiv -1+3\sqrt{2} \pmod{3^2}$. Cubing and...

What is the largest even integer that cannot be written as the sum of two odd composite numbers? Source: AIME 1984 , problem 14. Note $24 = 3\cdot 3 + 3\cdot 5$, and so if $2k$ has a representation as the sum of even multiples of 3 and 5, say $2k = e_3\cdot 3 + e_5\cdot 5$, we get a representation of $2k+24$ as a sum of odd composites via $2k+24 = (3+e_3)\cdot 3 + (5+e_5)\cdot 5$. But by the Frobenius coin problem every number $k > 3\cdot 5 -3-5 = 7$ has such a representation, hence every number $2k > 14$ has a representation as the sum of even multiples of 3 and 5. Thus every number $n > 24+14=38$ has a representation as the sum of odd composites. Checking, we see that $\boxed{38}$ has no representation as a sum of odd composites.