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# What Makes a Hard GMAT Question: Part 2

In part 2 of this series, we’ll look at another major criterion that determines the difficulty level of hard GMAT questions: **Disguising the Mathematical Concept the Question is Addressing.**

As in part 1, I suggest that you attempt both questions first, then read on to see my analysis.

Most test-takers don’t have difficulty re-learning the basic math addressed on the exam, but have difficulty identifying when/how to use these basic principles when the test-makers introduce these concepts in unorthodox situations.

**Case #1:**

Question #1:

If n = 20! + 17, then n is divisible by which of the following?

I. 15

II. 17

III. 19

(A) None

(B) I only

(C) II only

(D) I and II

(E) II and III

Question #2:

Does the integer k have a factor p such that 1<p<k?

(1) k > 4!

(2) 13!+2≤k≤13!+13

Analysis:

Both questions address the principles in number theory that: multiple of x + multiple of x = multiple of x and multiple of x + non-multiple of x = non-multiple of x. However, the construction of the first question (both its wording and the numbers as potential answers) make the need for implementing the above rule relatively straightforward. Since 20! contains 17 as a factor and 17 by definition contains itself as a factor, we know that 20! + 17 is an instance of *multiple of 17 + multiple of 17 = multiple of 17*. Thus, we know choice II is true. Using a similar logic, we know that 20! also contains 15 and 19 as factors, but that 17 does not, so we know that 20! + 17 is an instance of multiple of 15 + non-multiple of 15 = non-multiple of 15 and multiple of 19 + non-multiple of 19 = non-multiple of 19. We can thus conclude that the answer is C (II only).

In the second example, on the other hand, both the relative inscrutability of the question (it’s a disguised way of asking whether n is a prime number) and the unorthodox construction of statement 2 make identification and consequent application of the above principle much more challenging. You might not have seen it, but statement 2 addresses the identical concept that statement 1 addresses, but, as is the theme here, the concept is more disguised! What’s important to note in statement 2 is this: Every number that satisfies the provided range will be an example of the rule: multiple of x + multiple of x = multiple of x. Think about 13! + 3: we know that 13! contains 3 as a factor and is thus a multiple of 3, and we know that 3 is a multiple of itself, so 13! + 3 must be a multiple of 3. Since the numbers in the provided range only go up to 13! + 13, we know that every integer in that range will satisfy the property multiple of x + multiple of x = multiple of x. Thus, we can conclude that every number in that range will *not* be a prime number, meaning statement 2 is sufficient and the answer is B.

So, whereas, in question 77, a test taker needs only to properly implement the concept, in the second example, the test-taker needs to both identify and implement the concept. Fewer test-takers will be able to do so, making example #2 more difficult.

**Case #2:**

Question #1:

Which of the following equations is NOT equivalent to 10y^{2}=(x+2)(x−2) ?

(A) 30y^{2}=3x^{2}−12

(B) 20y^{2}=(2x−4)(x+2)

(C) 10y^{2}+4=x^{2}

(D) 5y^{2}=x^{2}−2

(E) y^{2}=(x^{2}−4)/10

Question #2:

If n = 3^{8} – 2^{8}, which of the following is NOT a factor of n?

(A) 97

(B) 65

(C) 35

(D) 13

(E) 5

Explanation:

In both examples, the algebraic concept of a difference of squares x^{2} – y^{2} = (x+y)(x-y) is fundamental to the proper problem-solving approach. However, test-takers mostly learn/implement this concept in situations where the respective terms are raised to a power of 2, so most-takers fail to see when those terms are raised to larger powers. So, whereas most test-takers familiar with these concept will be able to properly implement it on question #1 (for example, you might see that Choice C can be written as 10y2 = x2 – 4 = (x+2)(x-2) they’ll have a much harder time doing so on question #2, where you need to see that 3^{8} – 2^{8} = (3^{4})^{2} – (2^{4})^{2} = (3^{4} + 2^{4})(3^{4} – 2^{4}) = (97)(65).