If $a \equiv 4 (mod 13)$ and $b \equiv 9 (mod 13)$, then how can we find $0 \leq c \leq 12$ such that $c \equiv 9a (mod 13)$ and $ c \equiv 2a+3b (mod 13)$?
2026-03-26 04:36:43.1774499803
How to find new number congruent to another one with respect to some given congruences?
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We need to find the residue of $c$ modulo $13$ so because $c\equiv9a(\bmod13)$, $c\equiv9\cdot4\equiv36\equiv10(\bmod13)$ $c\equiv2a+3b\equiv2\cdot4+3\cdot9\equiv8+27\equiv35\equiv9(\bmod13)$ So we got that $c\equiv9\equiv10(\bmod13)$ which is obviously false, so we have no solutions.