Large Numbers Don't Sway Me!

From the first condition $30x \equiv 877 \mod p$ , we multiply both sides by $4$ so we have $120x \equiv 3508 \mod p$ . Similarly multiplying both sides of the second condition $24x \equiv 107 \mod p$ by $5$ gives $120x \equiv 535 \mod p$ and so subtracting the two equations give $0 \equiv 2973 \mod p$ . The factors of 2973 are 1, 3, 991 and 2973 so the only value of $p$ that is satisfied is $p = \boxed{991}$ .

To find the value of $x$ , we now subtract the two original equations to get $6x \equiv 770 \mod 991$ . If a solution exists to this then we can write $6x = 770 + 991k$ . Taking modulo 6 of this gives $0 \equiv 770 + 991k \equiv 2 + k \mod 6$ . Using the smallest positive value of $k$ that satisfies this (which is $k = 4$ , we have now $6x \equiv 770 + 991*4 = 4734 \mod991$ which implies $x \equiv 789 \mod 991$ and so we have $x = \boxed{789}$ .

Adding $x+p$ gives us $\boxed{1780}$ .