Wednesday, March 28, 2012

Putnam 1998 B5

Let $N$ be the positive integer with $1998$ decimal digits, all of them $1$; that is, \[ N = 1111 \cdots 11. \] Find the thousandth digit after the decimal point of $\sqrt{N}$.

Our first B-side is much easier than it looks. There are lots of ways to approach this problem, but let's think approximately first. What's the idea here? The decimal expansion for $1/9$ has a bunch of $1$'s in it, and $\sqrt{1/9} = 1/3$. Hmm. If we just move the decimal point to the left...So $N$ is about $1/9 \cdot 10^{1998}$, so $\sqrt{N}$ is about $1/3 \cdot 10^{999}$. Well, we obviously need a better approximation.

So $N$ actually equals $1/9 (10^{1998}-1)$, so let's call its square root $1/3 (10^{999} - x)$, where we expect $x$ to be something really small. The idea will be to get a handle on the size of $x$. Here's the equation: \[ \begin{align*} \left( \frac13 ( 10^{999} - x) \right)^2 &= \frac19 (10^{1998}-1) \\ \frac19 (10^{1998} - 2x10^{999} + x^2) &= \frac19 (10^{1998}-1) \\ 2x 10^{999} -x^2 &= 1 \end{align*} \] That suggests that $x$ is about $10^{-999}/2$ (the $x^2$ is negligible). So if we had to guess at gunpoint right now, we'd say that the square root is about $\frac13 ( 10^{999} - 10^{-999}/2 )$, which in decimal expansion turns out to be $999$ 3's in front of the decimal point, then $999$ 3's to the right of it, then a 1, then a bunch of 6's. (Maybe it's easier to see the pattern if you replace $1998$ and $999$ by $2$ and $1$, $4$ and $2$, etc, so e.g. $\sqrt{11}$ is about $3.316$, $\sqrt{1111}$ is about $33.33166$, etc.) So our guess would be that the answer is $1$, and the rest of the problem is just about proving that our guess is right.

Let's try this: let $a = 1/3 (10^{999} - 10^{-999}/2)$, and see what $a-\sqrt{N}$ is. Well, $a^2-N = 10^{-1998}/36$, which is tiny, so $(a-\sqrt{N})(a+\sqrt{N}) = 10^{-1998}/36$, so $a-\sqrt{N} = 10^{-1998}/(36(a+\sqrt{N})) < 10^{-1998}$, so our approximation of $\sqrt{N}$ by $a$ should be good to at least $1998$ places to the right of the decimal point. That's plenty.

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