"Left-over" factors in division?

Hello!

Without having the actual problem in front of me I think they mean that 5 is the only factor of 210 that does not get cancelled out when we divide by 42. In other words 210 ÷ 42 is equal to 5:

210/42 = (7 × 3 × 2 × 5)/(7 × 3 × 2) = 5

And so the only factors of this quotient are 5 and 1.

I agree that the term "left over" is confusing, since you could easily mistake that for an indication of a "remainder" problem when in fact it has nothing to do with remainders.

akovalev wrote:MGMAT Advanced Quant guide, Chapter 1, In-Action Problem 1. The explanation on page 46 says: "We can look at the 'left-over' factors when we divide 210 by 42, of which there is only one prime (5), and of course, 1". What do they mean by "left-over" factors--what's left of the prime factorization of 210 when taking out the prime factorization of 42 (and 1 is considered in addition to the "left-overs")?

Pls post the question!

Sorry for the delay (just took the test yesterday evening), but here is the question:

Each factor of 210 is inscribed on its own plastic ball, and all of the balls are placed in a jar. If a ball is randomly selected from the jar, what is the probability that the ball is inscribed with a multiple of 42?

A. 1/16
B. 5/42
C. 1/8
D. 3/16
E. 1/4