It’s amazing how something as simple as a (!) symbol can throw us all for a loop, even when really we’re just looking at a very simple divisibility question. This is my own question designed to mimic a GMATPrep question.
Start by setting a timer for 2-minutes and try this one on your own, then scroll down for the explanation!
Which of the following is an integer?
I. 10! / 3!
II. 9! / 8!
III. 13! / 11! 5!
A) I only
B) II only
C) III only
D) I and II only
E) I, II, and III
If the result of a fraction is an integer that means that what is in the denominator divides evenly into what is in the numerator. The question for each of these Roman numerals becomes: will the denominator go evenly into the numerator?
Since we know that 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, and 3! = 3 x 2 x 1, it is obvious that the denominator of Roman Numeral 1 will go evenly into its numerator.
This same logic can be applied with Roman Numeral 2. As long as the factorial in the numerator is larger than the factorial in the denominator, then we will get a resulting integer.
Roman Numeral 3 is more complex. If we start by cancelling out the 11! from both the numerator and the denominator, the resulting fraction becomes:
(13 x 12) / (5 x 4 x 3 x 2)
Since 4 x 3 = 12, we can cancel those values out:
13 / (5 x 2)
Now we are stuck. 13 is a prime number, so neither 5 nor 2 will divide evenly into it. The result will be a decimal.
The correct answer is (D).