Answer:
234
Step-by-step explanation:
By Euclidean Algorithm on 13 and 18, we found that 13(7)+18(−5)=1.
Then we multiply the equation by n, we got
13(7n)+18(−5n)=n.
Thus 13(7n)+18(−5n)=n always has integer solution.
However, a and b must be positive integers. So we need to reduce 7n and increase -5n
to make them positive. Noting that13(18m)=18(13m)∀m∈Z.
⇒13(7n−18m)+18(−5n+13m)=n⋯(1)
Let a=7n−18m and b=−5n+13m.Since a and b are positive integers, we have two inequalities:
7n−18m>0 and −5n+13m>0
⇒5n13<m<7n18⋯(2)
In order to make 13a+18b=n has no solutions,n must be taken to make (2) is impossible.
That means there exists no such integer m lying in the interval (5n13,7n18).
As a consequence, the difference between
5n13 and 7n18 must not greater than 1.
7n18−5n13≤1⇒n13×18≤1⇒n≤234.
Let’s put the largest candidate n=234 into (2), we get 90<m<91 ⇒ m is not an integer, which is a contradiction!
Therefore the largest integer n to make 13a+18b=n has no positive integer solutions is 234.
Original Link: https://www.quora.com/If-a-and-b-must-be-positive-integers-what-is-the-largest-integer-n-such-that-13a-18b-n-has-no-solutions
