A two-digit number is equal to the sum of the digits and thrice the product of the digits.

Let the 2-digit number be 10a+b ( a and b are the tens and unit digit respectively)
Now based on the question we can write the equation as 10a+b = a +b+ 3(a*b) --Eqn1
10a+b = a+b+3ab --> 9a=3ab , Divide both sides by 3 --> 3a=ab --Eqn2 --> 3a-ab=0 --> a(3-b)=0 , now either a=0 or 3-b=0 , we are already given that the number is a 2-digit number so a cannot be 0 , thus 3-b=0 --> b=3 , alternatively , we could have divided by a on both sides of Eqn2 to get 3=b ( since a is positive we can divide)
Thus the number would be 10a +3 or a3 i.e. it can be 13,23,33,43,53,63,73,83,93 etc..

To be double sure we can test the Eqn1 also , lets say the number is 23 , so a= 2 and b = 3 , thus based on the Eqn 1 we can write 10*2 + 3 = 2+3+3(2*3) --> 20 +3 = 5 + 3*6 --> 23 = 5 + 18 --> 23 = 23 , thus LHS = RHS and thus the series of numbers we found is correct .
Thus since we can have different series of numbers less than or greater than or equal to 83 , ans D