Largest three digit common term in two APs

We have $a_n = 6n + 7$ and $b_m = 7m + 8$.

Let $X$ be a term common to both sequences: $X \equiv 7 \equiv 1 \pmod 6$ $X \equiv 8 \equiv 1 \pmod 7$

By Chinese Remainder Theorem, $X \equiv 1 \pmod{\text{LCM}(6, 7)} \implies X \equiv 1 \pmod{42}$.

So $X = 42k + 1$ for some integer $k$.

To find the largest 3-digit common integer, we set $X \le 999$: $42k + 1 \le 999 \implies 42k \le 998 \implies k \le 23.76$

Taking $k = 23$: $X = 42(23) + 1 = 966 + 1 = 967$