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Re: sqrts and primes (m06q31) [#permalink]
09 Aug 2010, 10:17

If , what is the value of integer ?

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

Re: sqrts and primes (m06q31) [#permalink]
10 Aug 2011, 09:51

1

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Let's solve it this way.

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.

Re: sqrts and primes (m06q31) [#permalink]
14 Aug 2013, 08:31

1

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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

Not Sufficient.

2) \sqrt{x+1} -1 = Prime \sqrt{x+1} -1 = P \sqrt{x+1} = P + 1 x+1 = P^2+2p+1 x = P^2+2p

if P=2, then x=8 If P=3, then x=15 If P=5, then x=35 If P=7, then x=63

so x = 8, 15, or 35

Not Sufficient.

1+2) X = 15 or 35 and then double check that each of 15 and 35 are divisible by two primes greater than 2.