Question
Solve $30u+26 \equiv 3\,\mod 7$
I know the solution is here,but i want to know what is flaw in my approach.
My Approach
$30u+26 \equiv 3\, \mod 7$
$30u \equiv -23\,\mod 7$
$u \equiv -23\times30^{-1}\, \mod 7 \tag{1}$
let's us find $30^{-1}\, \mod 7$
$30=7\times4+2$
$7=2\times3+1$
$1=7-2\times3$
$1=7-3\times(30-7\times4)$
$1=13\times7-3\times30-$
$30^{-1}\,\mod 7=3$
putting the value of $30^{-1}\,\mod 7$ in $(1)$,
$\endgroup$ 4$u \equiv -23\times3\, \mod 7$
$u \equiv -69\,\mod 7=1$but answer is $6$. Where am I wrong?
3 Answers
$\begingroup$value of $30^{-1}\text{mod} \,7=-3$ which i have written $3$.
Hence putting the value of $30^{-1}\text{mod}\, 7$ in $(1)$,
$u \equiv -23*(-3) \text{mod}\,7$
$\endgroup$ 4 $\begingroup$$u \equiv 69 \text{mod}\,7=6$
$$30n+26-3\equiv2n+2\equiv2(n+1).$$ Thus, $$n\equiv-1(\operatorname{mod}7)$$ or $$n\equiv6(\operatorname{mod}7)$$
$\endgroup$ 1 $\begingroup$When you want to solve a linear modulo n problem, say to yourself two things:
$\quad$ I can't stand seeing negative signs.
$\quad$ I can't stand seeing numbers $\ge n$.
A chain of equivalent equations:
$30u+26 \equiv 3\, \mod 7$
$2u+5 \equiv 3\, \mod 7$
$2u \equiv 3 -5\, \mod 7$
$2u \equiv -2\, \mod 7$
$2u \equiv 5\, \mod 7$
$u \equiv (2^{-1}) (5)\, \mod 7$
$u \equiv (4) (5)\, \mod 7$
$u \equiv 20\, \mod 7$
$u \equiv 6\, \mod 7$
