Manhattan GMAT Challenge Problem of the Week – 25 May 2010

by Manhattan Prep, May 25, 2010

Here's the latest Challenge Problem! As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go!

Question:

The smallest prime factor of 899 is x. Which of the following is true of x?

a. 1 < x 7

b. 7 < x 14

c. 14 < x 21

d. 21 < x 28

e. 28 < x 35

Answer:

One path to the solution involves brute force. We can test primes in order of size, applying divisibility rules that we know for small numbers, such as 3. However, all the simple rules fail. This method may wind up being the quickest way, but it is laborious.

The shortcut in this problem involves wishful thinking. 899 is awfully close to a nice number: 900. The reason 900 is so nice is that it is a square: [pmath]30^2[/pmath] = 900. (By the way, since we know from the wording of the problem that 899 has a prime factor less than itself, at least one of the prime factors must be below the square root of 899, and at least one prime factor must be larger than the square root of 899. This square root is just under 30. This is another reason why we might think of the nearby perfect square, 900.)

So we can write 899 = 900 1 = [pmath]30^2[/pmath] 1.

Now, ideally we would notice that we can take one step further and rewrite [pmath]30^2[/pmath] 1 as [pmath]30^2[/pmath] [pmath]1^2[/pmath], since 1 = [pmath]1^2[/pmath]. Why would we do this? Because now we have written 899 as a difference of squares, which we should know how to factor:

899 = 900 1 = [pmath]30^2[/pmath] 1 = [pmath]30^2[/pmath] [pmath]1^2[/pmath] = (30 + 1)(30 1) = 31 29.

Both 31 and 29 are prime numbers. The smallest prime factor of 899, therefore, is 29.

The correct answer is E.

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