# Puzzles

## Lots of ones

Is any of the numbers 11, 111, 1111, 11111, ... a square number?

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No. If one of them were a square number, then its square root must end in 1 or 9 (as this is the only way to make the final digit a one). So the square root is of the form \(10n\pm1\).

$$111...1=(10n\pm1)^2$$$$=100n^2\pm20n+1$$
$$=10(10n^2\pm2n)+1$$

If \(10(10n^2\pm2n)+1\) is of the form 111...1, then \(10n^2\pm2n\) is also of the form 111...1 (as it has just had the final 1 taken off). But \(10n^2\pm2n\) is even and 111...1 is odd, so this is not possible.

## What is the sum?

What is \(\displaystyle\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}\)?

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Start by rationalising the denominators:

$$\begin{array}{rl}
\frac{1}{\sqrt{n}+\sqrt{n+1}}
&=\frac{1}{\sqrt{n}+\sqrt{n+1}}\times\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}-\sqrt{n+1}}\\
&=\frac{\sqrt{n}-\sqrt{n+1}}{n-(n+1)}\\
&=\sqrt{n+1}-\sqrt{n}\end{array}$$

Therefore:

$$\begin{array}{rl}
\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}
&=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+\\&...+(\sqrt{15}-\sqrt{14})+(\sqrt{16}-\sqrt{15})\\
&=-\sqrt{1}+\sqrt{16}\\
&=3
\end{array}
$$