GMAT Tip: GMAT Math Strategy – Think with your pen
Today’s GMAT Tip comes from our friends at Knewton. In this post, they provide helpful hints on how to tackle particularly complex math problems. Read on to see what they have to say!
Here is a rather challenging GMAT math problem. Give it a shot:
For every positive EVEN integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is
(A) between 2 and 10
(B) between 10 and 20
(C) between 20 and 30
(D) between 30 and 40
(E) greater than 40
So, where do you start?
It is difficult to figure out how to approach a question like this one, since it’s a mix of so many different concepts: properties of even numbers, prime numbers, factors, and functions. No one wants to see a complex question like this on the GMAT, especially when you’re struggling to keep up your pace and finish the section on time.
The best advice for tackling complicated questions like this? Don’t waste too much time thinking about the perfect way to start. Just start writing!
Start by writing out what you know:
n is EVEN
Function h(n) = product of even integers from 2 to n
h(100) + 1
p is the smallest factor of h(100) + 1
Once you take the important elements out of the word problem, the problems should become clearer. The most important value in the problem is h(100) + 1, so let’s try to find out what that is.
h(n) = product of even integers from 2 to n.
So h(100) = product of even integers from 2 to 100.
h (100) = 2 x 4 x 6 x 8 … x 100
Hmm… now what? We don’t have time to calculate the value of h(100), so let’s see what we can do with the expression. Since the question is asking about prime factors of h(100) + 1, it will probably be helpful to find the prime factors of h(100). Let’s start there.
We know that h(100) is the product of only even integers, so let’s factor out a 2 from each number.
h(100) = 2 x 4 x 6 x 8 …. x 100 = 2(1) x 2(2) x 2(3) x 2(4) … 2(50)
Factoring out all the 2s (and there are 50 of them!) gives us:
2^50(1 x 2 x 3 x 4 … x 50)
We just found all the factors of h(100)! The factors are 2, and every integer from 1 to 50 inclusive. But the question is asking us about h(100) + 1. Now what?
Okay, let’s think about it. This is a good place to test the function with simpler numbers. If I know 3 and 4 are factors of 12, are they also factors of 12 + 1? No. 12 + 1 is 13, which is a prime number. So essentially, if x is a factor of n, then x will not be a factor of n+1, since n+1 cannot be a multiple of x. Test it out with other values if you’re not sure.
And since the integers from 1 to 50 are all factors of h(100), none of them can be a factor of h(100) + 1! Ah ha!
If p is the smallest prime factor of h(100) + 1, then p must be greater than 50. So the answer must be E, greater than 40.
Whew! In this problem, like in many others, just getting started allowed us to see more clearly what we were dealing with. If we had spent all our time trying to figure out how to start, we would have been out of time before we knew it.
Takeaway: When in doubt about how to solve a GMAT math problem, start writing, and then look for the most important value to find or solve.
Start practicing thinking with your pen!
For more information on Knewton, download Clear Admit’s independent guide to the leading test preparation companies here. This FREE guide includes coupons for discounts on test prep services at ten different firms!