WEEK'S PUZZLE 9

PUZZLE 9 – from the 30th November to 6th December( to be collected by the 7th of December)
Try to find all the two-digit numbers which, when divided by the sum of their digits, have a quotient equal to 7 with no remainder.


SOLUTION
The number in polynomial form is 10a+b. Because of the condition:
(10a+b)/(a+b) = 7
10a+b = 7a + 7b
3a = 6b
a = 2b
With b= 1,2,3,4 (more it's impossible), the solutions are: 21, 42, 63, 84