lean_workbook_plus_64311

As in the first solution, we start by considering the expression modulo $7$ . It becomes $x^3 + 2x^2 + 5x+3 \equiv 0 \pmod 7$ . Checking integers $0,1,2,3,4,5,6,7$ , we see that this is only satisfied for $x\equiv 3,4,5 \pmod 7$ . Now, considering the expression modulo $11$ , we get $x^3 + 8x^2 + 2x \equiv 0 \pmod {11}$ . Checking the possible residues, this is satisfied for $x \equiv 0 , 1 , 2 \pmod {11}$ . Finally, modulo $13$ , we get $x^3 + 4 x^2 -11 \equiv 0 \pmod {13}$ , which is satisfied when $x\equiv 2,3,4 \pmod {13}$ . Now, we can check integers $2,3,4 \pmod {13}$ until one works. They are $2,3,4,15,16,17,28,29,30,41,42,43,54,55,56,67$ . Now, $67 \equiv 4 \pmod 7, 67 \equiv 2 \pmod {13}$ , and $67 \equiv 1 \pmod{11}$ . Thus, our answer is $\boxed{67}$ (and you can check to see it works )$.