[ Residue arithmetic (other).]

Solve the congruences:
a) $8 x \equiv 3(\bmod 13)$;

b) $17 x \equiv 2(\bmod 37)$;

c) $7 x \equiv 2(\bmod 11)$

d) $80 x \equiv 17(\bmod 169)$.
a) $8 x \equiv 16(\bmod 13)$, so $x \equiv 2(\bmod 13)$.

b) First method. $-20 x \equiv 2(\bmod 37) \Rightarrow 10 x \equiv 36(\bmod 37) \Rightarrow 5 x \equiv 18 \equiv 55(\bmod 37) \Rightarrow x \equiv 11(\bmod 37)$.

Second method. Solve the equation $17 x-37 y=2$ in integers.

$17(x-2 y)-3 y=2$; by making the substitution $z=x-2 y$, we get $3(5 z-y)=2(1-z)$. From here $1-x=3 t, 5 z-y=2 t, z=1$ $-3 t, y=5 z-2 t=5-17 x$

$x=2 y+z=11-37 t$.

c) $7 x \equiv 35(\bmod 11)$, so $x \equiv 5(\bmod 11)$.

d) First method. $80 x \equiv 186(\bmod 169) \Rightarrow 40 x \equiv 93 \equiv 262(\bmod 169) \Rightarrow 20 x \equiv 300(\bmod 169) \Rightarrow x \equiv 15(\bmod$ 169).

Second method. $80 x \equiv 17(\bmod 169) \Rightarrow 2 x \equiv 4(\bmod 13) \Rightarrow x \equiv 2(\bmod 13)$, that is $x=13 n+2.80(13 n+2) \equiv 17$ $(\bmod 169) \Rightarrow 80 n+11 \equiv 0(\bmod 13) \Rightarrow 2 n \equiv 2(\bmod 13) \Rightarrow n=13 k+1 \Rightarrow x \equiv 15(\bmod 169)$.

## Answer

a) $x \equiv 2(\bmod 13)$;
b) $x \equiv 11(\bmod 37)$
c) $x \equiv 5(\bmod 11) ; \quad d) x \equiv 15(\bmod 169)$.

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