ah i see, so if it was 64 for instance, then the answer would be
(2,64) (4,32) and (8,16)
and the values for x and y can be swapped around, as long as that pair is the same.. ?
I have edited it again (in red), to indicate what I believe the problem is actually asking for. That is, you want to determine the pairs of natural numbers {x,y} such that the expression mx+ny (where m and n are non-negative integers) fails to take exactly 65 values among the natural numbers.
There is a result stating that if x and y are coprime then the largest integer that cannot be expressed in the form mx+ny is xy–x–y. This result is apparently referred to by some people as the Chicken McNugget theorem. In more traditional circles, the number xy–x–y is known as the Frobenius number for the set {x,y}. An old result of Sylvester (quoted here) states that if x and y are coprime then the number of unattainable linear combinations of x and y is .
If we want that number to be 65 then, as you can easily see by listing all the possible factorisations of 130, the possible pairs {x,y} (with neither x nor y equal to 2) are {3,66}, {6,27} and {11,14}.
Yes, except that you need to remember that these factors give you the numbers x–1 and y–1. You have to add 1 to each factor to get x and y.
So if you are told that there are 64 unattainable numbers then the possible values of {x,y} (excluding the case when x or y is 2) are {3,65}, {5,33} and {9,17}. The braces (curly brackets) are meant to indicate that these are unordered pairs, in other words it doesn't matter which way round they are.
that theorem states the largest value which cannot be made is (x-1)(y-1) -1
however, when x=2 and y =131, this would give 129... however, 265 is also unattainable and so on... all numbers greater than 131 which are not multiples of it and are odd will also be unattainable??