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1. is divisible by at least 2 prime numbers greater than 2 2. is prime

From 1, x is under the form a*b*c with a and b prime numbers greater than 2, with c integer. Several possibilities, among which 15 (=3*5*1) and 35 (=7*5*1)

From 2

Let's write 2 as N=squareroot(x+1)-1 (a) N is a prime number

From statement 2, we know that x+1 under radical must be an integer b/c the whole number must be an integer (a prime). Right? So find all numbers for x for which radical x+1 would yield an integer. So, lets organize it this way:

(1) \(x=3*5=15<53, \, x=5*7=35<53.\) No need to look for numbers that have 3 or more distinct prime factors. Not sufficient.

(2) \(\sqrt{x +1} - 1=p,\) where \(p\) is some prime number. Then \(\sqrt{x +1} =p+ 1\) or \(x+1=(p+1)^2\), and finally \(x=p(p+2).\) We can have \(x=3*5=15<53,\, x=5*7=35<53,\) and that's it. \(7*9=63>53.\) But still, more than one possibility. Not sufficient.

(1) and (2) together won't help either, as can be seen from the above.

Answer E.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

I got totally lost on this one and looking forward to a correct, faster solution. Here's my process:

0<x<53, so x = ?

1) x is divisible by at least 2 prime numbers greater than 2 which quickly rephrases to 15<=x<=51 although there are some numbers in there that probably won't fit

(1) x is divisible by at least 2 prime numbers greater than 2 (2) \sqrt{x +1} - 1 is prime

Stmt 1: x is divisible by 3 or 5 or 7 or 11...so on. So x can be 15 (3X5) or 35 (5X7) NOT SUFFICIENT

Stmt 2: sqrt(x+1) - 1 = P (lets denote a prime number with P). (x+1) = (p + 1)^2 x+1 = p^2 + 1 + 2p x = p^2 + 2p

Okay, now start putting in primes. if P=2, x = 8 if p=3, x = 15

Two different values --> stmt2 NOT SUFFICIENT

(1)+(2): The values of x from stmt 2 are 8, 15, 24, 35, 48. So, X can still be 15 or 35. Therefore even together the statements are not sufficient. Hence E.

gmatclubot

Re: sqrts and primes (m06q31)
[#permalink]
30 Jul 2014, 05:01