Foundations, part 13

$n\equiv{24}\pmod{x} \implies x \mid n-24 \implies x \mid 2n-48$

$n\equiv11\pmod{x} \implies x \mid 2n-11$

We conclude that $x \mid (2n-11)-(2n-48) = 37$

$n > 24 \qquad \therefore \boxed{n = 37}$

We have that $N = 24 + aX$ and $2N = 11 + bX$ for some integers $a,b$ . Then

$2N - N = (11 + bX) - (24 + aX) \Longrightarrow N = -13 + (b - a)X = -13 + cX$ for some integer $c$ .

Thus $24 + aX = -13 + cX \Longrightarrow 37 = (c - a)X = dX$ for some integer $d$ . Since $37$ is prime and $X$ is (assumed) positive and greater than $24$ we must have that $X = \boxed{37}$ .

Consider 24 and 2 x 24 = 48. Obviously 24 satisfies the first restriction. Now choose 25 for X. Then 48 mod 25 = 23 mod 25. So for X = 24 + n, 2N is 24 - n mod X. Since 24 - n = 11 it gives n = 13 and thus X = 37.