jmc

Divide the first congruence by 7, remembering to divide 14 by $\text{gcf}(7,14)=7$ as well. We find that the first congruence is equivalent to $x\equiv 1(\mathrm{mod}2)$. Subtracting 13 from both sides and multiplying both sides by 5 (which is the modular inverse of 2, modulo 9) gives $x\equiv 6(\mathrm{mod}9)$ for the second congruence. Finally, adding $2x$ to both sides in the third congruence and multiplying by 17 (which is the modular inverse of 3, modulo 25) gives $x\equiv 17(\mathrm{mod}25)$. So we want to solve $$\begin{array}{rl}x& \equiv 1(\mathrm{mod}2)\\ x& \equiv 6(\mathrm{mod}9)\\ x& \equiv 17(\mathrm{mod}25).\end{array}$$Let's first find a simultaneous solution for the second and third congruences. We begin checking numbers which are 17 more than a multiple of 25, and we quickly find that 42 is congruent to 17 (mod 25) and 6 (mod 9). Since 42 does not satisfy the first congruence we look to the next solution $42+\text{lcm}(25,9)=267$. Now we have found a solution for the system, so we can appeal to the Chinese Remainder Theorem to conclude that the general solution of the system is $x\equiv 267(\mathrm{mod}450)$, where 450 is obtained by taking the least common multiple of 2, 9, and 25. So the least four-digit solution is $267+450(2)=\boxed{1167}$.