taim wrote:I know the answer to this problem, but I would like to know how to approach this question.
How & why does the solution in OG11 recognise that 441 = (9)(49) ?

Can you please write the question here?

taim wrote:I know the answer to this problem, but I would like to know how to approach this question.
How & why does the solution in OG11 recognise that 441 = (9)(49) ?

The question asks for the number of positive divisors of 441. To answer such a question, we'd normally want to prime factorize 441. If you need to prime factorize a large number on the GMAT, it will almost always be the case that 2, 3 or 5 is one of the factors - otherwise the question would be too time consuming. We can see that 2 is not a factor of 441 (since 441 is odd), and that 5 is not a factor (since 441 does not end in 0 or 5).

Now, there is a very useful trick that lets us identify multiples of 3 and of 9: if the sum of a number's digits is a multiple of 3, the number itself is a multiple of 3, and if the sum of a number's digits is a multiple of 9, the number itself is a multiple of 9 (note that this only works for 3 and 9 - it doesn't work for any other divisors!). Adding the digits of 441, we get 4+4+1 = 9, which is a multiple of 9, so 441 must be divisible by 9. Once you see that, you can find that 441 = (3^2)(7^2) using whatever method you like.

Finally, once you have the prime factorization of a number, you can count all of its divisors by adding one to each exponent and multiplying what you get. In the factorization (3^2)(7^2), our exponents are 2 and 2, so adding one to these numbers and multiplying, we find that 441 has (2+1)(2+1) = 3*3 = 9 positive divisors.