The Angry Statistician

Problem: Waldo is playing Asteroids. This game starts with 3 ships and you earn an extra ship for every 10000 points you score. At the end of his first game, Waldo noticed that he had the lowest score possible while averaging exactly 9000 points per ship. At the end of his second game, Waldo noticed that he had the highest score possible while averaging exactly 9000 points per ship. What were his two scores? Solution: Let the number of starting ships be $S$, points for an extra ship be $X$ and the player's average scoring rate be $A$. Furthermore, let $R = A/X$ and the number of bonus ships earned by the player during the game be $B$. The number of bonus ships earned by the player during his game is the floor of his final score over the bonus value $X$. Algebraically, we have $$B = \lfloor (S+B)\cdot R \rfloor.$$ Now let $$(S+B)\cdot R = \lfloor (S+B)\cdot R \rfloor + f,$$ where $0 \leq f < 1$. This gives us \begin{aligned} (S+B)\cdot R &= B...

Problem: Find the smallest positive number that doubles when you move the last digit to the front. Solution: The answer is $105263157894736842$, and the solution is similar to Twisty Temperature . Let this number be $$x = x_{n}\cdot 10^{n-1} + ... + x_{1}\cdot 10^1 + x_{0}$$ with $0 < x_{n} < 10$, then after moving the last digit to the front we get $$y = x_{0}\cdot 10^{n-1} + (x-x_{0})/10.$$ We also have that $y = 2 x$, so our equation becomes $$19\cdot x = x_0 \cdot (10^{n}-1).$$ Now 10 is a primitive root modulo 19, and so it follows that $x$ is integral if and only if $n$ is of the form $18\cdot m$. Also note that we need $2 \le x_0 \le 9$ since we require $0< x_{n} < 10$. When $m=1$ and $x_0 = 2$ we get $x  =105263157894736842$; when $m=2$ and $x_0 = 2$ we get $$2 (10^{36}-1)/19 = 105263157894736842105263157894736842.$$ That this number indeed doubles when you move the last digit to the front can be...

Problem: When Waldo recently did a conversion of a positive integral Celsius temperature $c = 275$ to its Fahrenheit equivalent $f$ (which turned out to be $527$ ), he noticed to his amazement that he could have simply moved the last digit of $c$ to the front to obtain $f$. Doing some intense calculations he failed to discover the next largest such example. Does one exist, and if so, what is it? Solution: Let $c = x_{n}\cdot 10^{n-1} + ... + x_{1}\cdot 10^1 + x_{0}$ with $x_{n} > 0$, then $$f = x_{0}\cdot 10^{n-1} + (c-x_{0})/10.$$ We also have that $$f = (9/5)\cdot c + 32.$$ Notice that in order for $f$ to be integral $c$ must be divisible by 5; this implies that $x_0=5$ since it cannot equal 0 (since as a number $f>c$). Our equation then becomes $$(9/5)\cdot c + 32 = 5\cdot 10^{n-1} + (c-5)/10$$ implies $$c = 5\cdot (10^n - 65)/17.$$ Now it turns out that 10 is a primitive root modulo 17, and so it follows that $c$ ...