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.