Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.3 National Olympiad Prove it Estonia

We say that two real numbers $r$ and $s$ are close if $|r - s| = 10^u$ for some integer $u$ . Let $y = ax + b$ be a linear function, for which there exist close numbers $x_1$ and $x_2$ so that the corresponding $y_1$ and $y_2$ are also close. Prove that for any close numbers $x'_1$ and $x'_2$ , the corresponding $y'_1$ and $y'_2$ are also close.

Solution

From the premises we get $|x_1 - x_2| = 10^u$ and $|y_1 - y_2| = |(ax_1 + b) - (ax_2 + b)| = 10^v$ for some integers $u, v$ . So,
$10^v = |(ax_1 + b) - (ax_2 + b)| = |a(x_1 - x_2)| = |a| \cdot |x_1 - x_2| = |a| \cdot 10^u,$
which gives $|a| = \frac{10^v}{10^u} = 10^{v-u}$ . Let $x'_1, x'_2$ be any close real numbers, $|x'_1 - x'_2| = 10^w$ . Then $|y'_1 - y'_2| = |(ax'_1 + b) - (ax'_2 + b)| = |a(x'_1 - x'_2)| =$

$|a| \cdot |x'_1 - x'_2| = 10^{v-u} \cdot 10^w = 10^{w+v-u}$ . Since $u, v, w$ are integers, $w + v - u$ is also an integer, which shows that $y'_1, y'_2$ are close.

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