Manhattan GMAT Challenge Problem of the Week – 11 Feb 10:

by on February 11th, 2010

Welcome back to this week’s 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 minutes and go!

Question

If n is a positive integer greater than 1, then p(n) represents the product of all the prime numbers less than or equal to n. The second smallest prime factor of p(12) + 11 is

(A) 2
(B) 11
(C) 13
(D) 17
(E) 211

Solution

The quantity p(12) equals the product of all the primes less than or equal to 12. Thus, the number we are looking for is this:

2×3×5×7×11 + 11

= 11×(2×3×5×7+1) [factor out the 11]
= 11×(210 + 1) [do the arithmetic]
= 11×211
= some large number.

We want to keep this number factored, and in fact we need to find its prime factorization. So, we ask: is 211 prime? Well, 211 cannot be divided evenly by 2, 3, 5, or 7, because 211 equals a multiple of all those numbers (210), plus 1. The “plus 1″ means that 211 won’t be divisible by any of the same factors as 210 (except for 1). Is 211 divisible by 11 or by 13? No, as we can check quickly by long division.

And we only need to check possible prime factors up to the square root of 211 (which is between 14, the square root of 196, and 15, the square root of 225). If there is a pair of factors besides 211 and 1, one of the factors in the pair must be lower than the square root of 211 (and the other factor in the pair would be larger). Since there are no prime factors of 211 less than 14, we know that 211 is prime, and the second smallest prime factor of p(12) + 11 = 11×211 is 211.

The correct answer is (E) 211.

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• Can someone please explain why 11 factored out?

• i have the same doubt. can someone answer harrison's question?

• Well, that's the whole theme of the question. The two parts to this sum P(12) + 11 have only 11 as the common factor. Hence you are expected to pull that 11 out. Once you pull that 11 out you have 2x3x5x7 + 1.

If it was some number plus 17, most of the time the question will revolve around you spotting that you can pull the 17 out of the two factors.

I have seen a lot of practice questions on this central theme. One plus of a certain multiple (2x3x5x7) will not have any of the factors of the multiple. For eg. 2x3x5 = 30. Add one to it and you get 31. 31 is not divisible by 2 or 3 or 5. Similar case with 211. I hope I cleared your doubt.

• Dear Vineesh, I totally agree with you about 211 as a prime.

But the point hereby is P(12) + 11= 1*11*211. So we should have 11 is the second smallest factor of P(12) + 1.

Any clearer explanation?

• Explanation:

1 is not a prime number (a prime number a positive integer that is divisible by 2 numbers: 1 and itself - excluding 1!!!)!

Therefore, we're left with 2 numbers: 11 and 211, 211 being the second smallest, and coincidentally the largest You could have left 11 inside, but then you would still have to final factor the result which would ultimately leads to 1x11x211.

Hope this helps.

• Nice responses, Vineesh and Jmx! You two are totally right. • nice explanation! I saw a similar problem on GMAT PREP 1!