Find, with proof, the smallest positive integer m 2 for which both of the following statements are true at the same time:

The equation 6x + 11y = m has no non-negative solutions.

For any k > m, the equation 6x + 11y = k has at least one non-negative solution.

The procedure of proof should follow the following format. The answer should like that.

Example

Proposition: For any a,b, if at least one of a,b is not zero, then GCD(a/GCD(a,b), b/GCD(a,b)) = 1

Proof. Let a and b be arbitrary integers.

(1)Since at least one of a,b is non-zero, by definition GCD(a,b) exists and is a positive integer.

(2)By Bzouts Lemma, there exist x and y such that ax + by = GCD(a,b)

(3)By (1), GCD(a,b) is a non-zero integer.

(4)From (2), ax + by = GCD(a,b), and by (3), we can divide both sides by GCD(a,b), so (a/GCD(a,b))x + (b/GCD(a,b))y = 1

(5)By the characterization of GCD, since 1 divides both a/GCD(a,b) and b/GCD(a,b), and from

(4) we know (a/GCD(a,b))x + (b/GCD(a,b))y = 1, it follows that GCD(a/GCD(a,b), b/GCD(a,b)) = 1.