An ordered pair of digits (a,b) is such that 4a5, b72 is a multiple of 66. Find a + b .
+33614 The prime factors of 66 are 2.3.11, so 4a5b72 must be divisible by these factors.
Clearly it is divisible by 2.
For it to be divisible by 3 we must have a+b as a multiple of 3.
Try the simplest: a = 0, b = 3: 405372/66 = 6142.
Assuming the sum of a and b is unique, then a+b = 3
