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