Question 1157079
<br>
Here is a solution using the method of Diophantine equations.<br>
A basic Diophantine equation is a single equation with two unknowns, in which the solutions are restricted to integers.  The requirement that the solutions be integers limits the solutions to only a few, or, as in this case, to a single solution.<br>
This example is a very common type, containing terms in m, n, and mn.  The basic techniques for solving that kind of equation are standard.<br>
(1) Solve the equation for one of the variables in terms of the other.<br>
{{{14mn = 55-7m-2n}}}
{{{14mn+7m = 55-2n}}}
{{{7m(2n+1) = 55-2n}}}
{{{7m = (55-2n)/(2n+1)}}}<br>
(2) Find the value(s) of n that make the fraction an integer.<br>
m has to be a positive integer, so 7m is an integer.  That means {{{(55-2n)/(2n+1)}}} is an integer.<br>
It is generally, as in this case, difficult to find values of n that make that expression an integer.  So this is what we usually do at this point: write the numerator of the fraction as a multiple of the denominator plus a constant.  It's as if we are doing the division and expressing the fraction as an integer plus a remainder.<br>
{{{7m = (56-1-2n)/(2n+1)}}}
{{{7m = (56-(1+2n))/(2n+1)}}}
{{{7m = 56/(2n+1)-1}}}<br>
Now 7m and 1 are integers, so {{{56/(2n+1)}}}must be an integer; and that means 2n+1 must be a divisor of 56.<br>
In general, we would look at all the factors of 56 and see what values of n would make {{{56/(2n+1)}}} an integer.<br>
But in this example we can go directly to the solution with a bit of logical reasoning.<br>
For all n, 2n+1 is odd; so 2n+1 must be an odd divisor of 56.<br>
But there is only one odd divisor of 56: 7.  (Okay, 1 is an odd divisor of 56; but that would make n=0, and the problem requires m and n to be positive integers.)<br>
So 2n+1 = 7, which makes n=3.<br>
Then {{{7m = 56/7-1 = 8-1 = 7}}} --> m = 1.<br>
ANSWER: (m,n) = (1,3)<br>
CHECK:
14mn = 14*1*3 = 42
55-7m-2n = 55-7-6 = 42<br>