An ordered pair of digits (a,b) is such that 4a5, b72 is a multiple of 66. Find a + b .
+2666 For a number to be divisible by 66, it has to be divisible by all of its prime factors, which are 2, 3, and 11.
The number is already divisible by 2.
For the number to be divisible by 3, \(a+b+18\) must equal a multiple of 3. 18 is already a multiple of 3, so \(a+b\) must equal
For the number to be divisible by 11, \(4-a+5-b+7-2\) must be a multiple of 11.
Simplifying the equation, \(14 - (a+b)\) must be a multiple of 11.
The only way we can do this is if \(a+b = 3\), because 11 is a multiple of 11, and 3 is a multiple of 3.
Thus, \(a + b = \color{brown}\boxed{3}\)
