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The modular arithmetic system ( $n \bmod 3, n \bmod 5$ ) for number 13 has the strange property that is actually (1,3). Explain how can all numbers with such depiction can be found, without finding 15 separate pairs of modulos, but solving the following equations.
$10x + y = x \pmod 3$
$10x + y = y \pmod 5$
Consider known that:
$10u + 6v = u \pmod 3$
$10u + 6v = v \pmod 5$

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In terms of pure modular equivalence:

$10x+y \equiv x \bmod 3$ requires $y\equiv 0\bmod 3$ ie single digit $y \in \{0,3,6,9\}$

$10x+y \equiv y \bmod 5$ is always true, regardless of the value of $x$. For a valid number, we need $x>0$.

However, considering $a\bmod b$ as a function that produces a result in the range $[\ \!0,b{-}1\ \!]$, we would be limited to $x\in\{1,2\}$ and $y\in\{0,3\}$, giving answers of $\{10,13,20,23\}$.

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