Answer:
31
Explanation:
We can solve the given equation for 'b', then find the integer values of 'a' that make 'b' a positive integer. There are 3 such values. One of these minimizes the objective function.
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solve for b
ab +5b = 373 +6a . . . . . . isolate b terms by adding 6a
b = (6a +373)/(a +5) . . . . . divide by the coefficient of b
b = 6 +343/(a +5) . . . . . . . find quotient and remainder
integer solutions
The value of 'b' will only be an integer when (a+5) is a factor of 343. The divisors of 343 = 7³ are {1, 7, 49, 343}. so these are the possible values of a+5. Since a > 0, we must eliminate a+5=1. That leaves ...
a = {7, 49, 343} -5 = {2, 44, 338}.
Possible values of b are ...
b = 6 +343/{7, 49, 343} = 6 +{49, 7, 1} = {55, 13, 7}
Then possible (a, b) pairs are ...
(a, b) = {(2, 55), (44, 13), (338, 7)}
objective function
The values of the objective function for these pairs are ...
|a -b| = |2 -55| = 53
|a -b| = |44 -13| = 31 . . . . . the minimum value of the objective function
|a -b| = |338 -7| = 331