I am trying to solve $$1151x \equiv 1 \pmod{903}$$ To do this, first I used the Euclidean algorithm: \begin{align} 1151 &= 1 \cdot 903 + 248 \\ 903 &= 3 \cdot 248+159 \\ 248 &= 1 \cdot 159 + 89 \\ 159 &= 1 \cdot 89 + 70 \\ 89 &= 1 \cdot 70 + 19 \\ 70 &= 3 \cdot 19 + 13 \\ 19 &= 1 \cdot 13 + 6 \\ 13 & = 2 \cdot 6+1 \end{align}
Where can I go from here?
You work backwards thusly:$$(A)...2( 6)=13-1\implies 2 (19)=2(13+6)=2(13)+(13-1)=$$ $$=3(13)-1\implies 2(19)+1=3(13).$$ $$(B)... 2(19)+1=3(13)\implies 3(70)=3((3)19+13)=$$ $$=9(19)+(2(19)+1)= 11(19)+1\implies 3(70)-1=11(19).$$ $$(C)... 3(70)-1=11(19)\implies 11(89)=11(70 + 19)=$$ $$=11(70)+(3(70)-1)=14(70)-1\implies 11(89)+1=14(70).$$ $$(D)... 11(89)+1=14(70)\implies 14(159)=14(89 +70)=$$ $$=14(89)+(11(89)+1)=25(89)+1\implies 14(159)-1=25(89).$$ $$(E)...14(159)-1=25(89)\implies 25(248)=25 (159+89)=$$ $$=25(159)+(14(159)-1)=39(159)-1\implies 25(248)+1=39(159).$$ $$(F)...25(248)+1=39(159)\implies 39(903)=39(3(248)+159)=$$ $$=117(248)+(25(248)+1)=142(248)+1\implies 39(903)-1=142(248).$$ $$(G)...39(903)-1=142(248)\implies 142(1151)=142(903+248)=$$ $$=142(903)+(39(903)-1)=181(903)-1\implies$$ $$\implies 142(1151)\equiv -1 \pmod {903} \implies$$ $$ \implies -142(1151)\equiv 1 \pmod {903} \implies$$ $$\implies 761(1151)=(903-142)(1151)\equiv 1 \pmod {903}.$$
All of the solutions of $1151x\equiv 1\pmod {903}$ are of the form $x=903y+761.$