Math League #5
Some problems from the Wisconsin Math Leauge. I scored 6/6! :D Riju and I agreed that it was probably the easiest one so far this year, though the last one still requires some thinking. I waited 24 hours to post this, and I hope that is a good grace period for the problems.
Problem 3 What $n \in \mathbb{Z}$ satisfies
\[n < \frac{1}{\log_{\frac{1}{2}}\frac{1}{3}} + \frac{1}{\log_{\frac{1}{5}}\frac{1}{3}} < n+1 ?\]
While this is really just a change in base exercise and then plugging it into a calculator for most people, we can still bound it without a calculator and get the answer. I’ll assume you have gotten to this point,
\[n < \frac{\log{\frac{1}{10}}}{\log{\frac{1}{3}}} < n+1.\]Set the base equal to $\frac{1}{3}$ to make the bottom part dissapear:
\[n < \log_{\frac{1}{3}}{\frac{1}{10}} <n+1.\]Past this all you need to know is that $\log_{\frac{1}{3}}{\frac{1}{10}}$ is somewhere between $\log_{\frac{1}{3}}{\frac{1}{9}} =2$ and $\log_{\frac{1}{3}}{\frac{1}{27}} =3$, so $n=2$. We’re done.
Problem 6 Which rational numbers $r$ give the quadratic
\[rx^{2} + (r-2)x+(r+2) = 0\]two integer roots?
We start by letting the two integer roots be $p,q$. After expanding $(x-p)(x-q)$, we find that
\[x^2 + (-q-p)x +qp = 0.\]We make the coefficient of the first equation $1$ by dividing it by $r$
\[x^2 + \frac{r-2}{r}x + \frac{r+2}{r} = 0.\]So we want the coefficents of these two equations to be the same, yeilding
\[-q-p = \frac{r-2}{r} \qquad\qquad qp = \frac{r+2}{r}\]The key point here is that we can solve both equations for $2$, hopefully removing the constants.
\[(p+q+1)r = 2 \qquad\qquad (qp-1)r = 2.\]Then finally setting these equal to each other, and dividing $r$, which is indeed possible because $r\neq 0 $:
\[p+q+1 = qp-1.\]The $p$, $q$ and $qp$ along with the fact that we are finding integer solutions screamed Simon’s favorite factoring trick.
\[\begin{align*} 0 &= qp - p - q - 2 \\ 0 &= (q-1)(p-1)-1-2 \\ 3 & = (q-1)(p-1). \end{align*}\]The solution is so close… we notice that the only way to get $3$ with integers is $(1,3),(3,1),(-1,-3)(-3,-1)$ (the last few negative ones almost messed me up because I missed them). Also note that the switching the pair does not do anything, as $q$ and $p$ will just swap, which has no affect on our value of $r$.
| $p-1$ | $q-1$ |
|---|---|
| $1$ | $3$ |
| $-1$ | $-3$ |
Solving for $p$, and $q$ (which I will leave as an exercise for the reader), we plug each value back into $ (p+q+1)r = 2 $ from before to get our two values of $r$: $\boxed{-2, \frac{2}{7}}$.
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