Manhattan GMAT Challenge Problem of the Week – 1 Feb 2011

by Manhattan Prep, Feb 1, 2011

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Question

Two prime numbers are considered consecutive if no other prime lies between them on the number line. If [pmath]p_1[/pmath] and [pmath]p_2[/pmath] are consecutive primes, with |[pmath]p_1[/pmath] [pmath]p_2[/pmath]| > 2, what is the smallest possible absolute value of the coefficient of the x term in the distributed form of the expression (x [pmath]p_1[/pmath])(x [pmath]p_2[/pmath])?

(A) 5

(B) 8

(C) 12

(D) 18

(E) 24

Answer

The question seems forbidding, but start by grabbing onto the most concrete part, which comes at the end. Start with distributing (x [pmath]p_1[/pmath])(x [pmath]p_2[/pmath]).

(x [pmath]p_1[/pmath])(x [pmath]p_2[/pmath]) = [pmath]x^2[/pmath] ([pmath]p_1[/pmath] + [pmath]p_2[/pmath])x + [pmath]p_1[/pmath][pmath]p_2[/pmath]

All we care about is the coefficient of the x term, which is ([pmath]p_1[/pmath] + [pmath]p_2[/pmath]). Specifically, we care about the absolute value of this, which is [pmath]p_1[/pmath] + [pmath]p_2[/pmath], since primes are by definition positive.

So what we are really asked for is the smallest possible value of [pmath]p_1[/pmath] + [pmath]p_2[/pmath], under two conditions:

1. These two primes are consecutive, meaning that theres no other prime between them.
2. |[pmath]p_1[/pmath] [pmath]p_2[/pmath]| > 2, meaning that the primes are more than 2 units apart on the number line.

In other words, the question really is what is the smallest possible sum of two consecutive primes that are more than 2 units apart?

Now take the first several primes: 2, 3, 5, 7, 11, 13. The first pair of consecutive primes more than 2 units apart is {7, 11}. Their sum is 18.

The correct answer is D.

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