Question 1210504
<br>
{{{2a+3b=100}}}<br>
This is a linear Diophantine equation -- one equation with two variables, with a finite number of solutions because the solutions are positive integers.  A common start on finding the set of solutions is to solve the equation for one variable.<br>
Because the numbers 2 and 100 are both even, it is probably fastest to solve the given equation for b.<br>
{{{3b=100-2a}}}
{{{b=(100-2a)/3}}}
{{{b=2(50-a)/3}}}
{{{b=2((50-a)/3)}}} [1]<br>
With the equation in the last form [1], we see that (50-a) must be divisible by 3.  That means, among other things, that consecutive values of a that provide solutions will differ by 3.<br>
(1) Solution with smallest value of a...<br>
Remembering that the solutions must have both a and b positive 2-digit integers, we can see that the smallest 2-digit value for a that makes (50-a) divisible by 3 is 11.<br>
That gives us {{{b=2((50-11)/3)=2(13)=26}}}
(2) Solution with the largest value of a...<br>
The solution with the largest value of a is the one with the smallest value of b.  Again since a and b must be 2-digit integers, equation [1] says that all values of b that give solutions will be even, so the smallest value of b that will give a solution is the smallest 2-digit even integer, which is 10.<br>
From (1) and (2), the solutions are the ones with b having even values from 10 to 26 inclusive.  The number of such solutions is<br>
{{{(26-10)/2+1=9}}}<br>
ANSWER: There are 9 solutions having both a and b 2-digit positive integers<br>