Find all two digit positive integers in which the difference between the integer and the product of its two digits is 12

Let ab be a 2 digit number, then ab=10a+b, where a and b are digits, a≠0.

"the difference between the integer and the product of its two digits is 12"

10a+b - (a*b)=12

10a+b-ab=12

factorize b:

10a+b(1-a)=12

subtract 10 from both sides, to produce a factor (1-a) or (a-1) and then factorize:

10a-10+b(1-a)=2

10(a-1)+b(1-a)=2

(a-1)(10-b)=2

(a-1)(10-b) can be (1,2), (2, 1), (-1, -2), (-2, -1)

if a-1=1 and 10-b=2, then a=2, b=8
if a-1=2, and 10-b=1, then a=3, b=9
if a-1=-1, then a=0, which is not possible
if a-1=-2, then a=-1, which is not possible

so the solutions (a, b) are (2, 8) and (3, 9)

Then the integers are 28, 39

28, 39

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.