How many five-digit numbers of the form 32x5y are divisible by 33?
For a number to be divisible by 33, it must be divisible by its prime factors, which, in this case, are 3 and 11.
For a number to be divisible by 3, its digits must sum to a multiple of 3. This means \(3+2+5+x+y=\) multiple of 3. Thus, \(x+y\) is 2,5,8,11,14, or 17.
For a number to be divisible by 11, the alternating sum of the digits must be divisible by 11. This means that \(3-2+x-5+y=\) multiple of 11. Thus, \(x+y=\) 4 or 15.
Now, we have to find a pair of numbers that met both these constraints.
Can you take it from here?