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sqrts and primes (m06q31) [#permalink]
 12 Nov 2007, 11:50
Question Stats:
71% (01:49) correct
28% (01:18) wrong based on 1 sessions
If 0 \lt x \lt 53 , what is the value of integer x ? 1. x is divisible by at least 2 prime numbers greater than 2 2. \sqrt{x +1} - 1 is prime Source: GMAT Club Tests - hardest GMAT questions Please show your work especially for 2. thanks
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stmt 1. X is divisible by at least 2 prime numbers greater than 2:
ex. 15,21,33,35,39,51,55,57,...195...etc
Insuff
stmt 2. sqrt(x+1) - 1 is prime
ex. 8,15,35,64...195
Insuff
Together
15,35,195
Insuff
Ans E
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Re: sqrts and primes [#permalink]
12 Nov 2007, 22:01
bmwhype2 wrote: What is the value of integer X if 0 < x < 53?
1. X is divisible by at least 2 prime numbers greater than 2
2. sqrt(x+1) - 1 is prime
Please show your work especially for 2. thanks
Get E.
Stat 1: x could be 15 or 21. Insuff.
Stat 2: x could be 15 or 8. Insuff.
Together: x could be 15 or 35. Insuff.
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Re: sqrts and primes (m06q31) [#permalink]
09 Aug 2010, 05:45
E 
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Re: sqrts and primes (m06q31) [#permalink]
09 Aug 2010, 08:15
2
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From 1 :
x is divisible by atleast 2 prime numbers greater than 2
so x can be 15 , 21 , 33, 35, 39 ...
Insuff
From 2 :
sqrt(x+1) - 1 is prime
=> sqrt(x+1) is even
Squaring both sides
x+1 is even
=> x is Odd
Still Insuff
Taking together
We don't get any unique value.. So answer is E
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Re: sqrts and primes (m06q31) [#permalink]
09 Aug 2010, 09:35
Let say M is 35 , which is greater than 0 and less than 53
1. Which is divisible by 5 and 7 which are 2 prime number greater than 2 .
2. Squ( M + 1 ) -1 => squ ( 35 + 1) –1 => 6 -1 => 5 which is again a prime number .
So I think with together it is sufficient .
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Re: sqrts and primes (m06q31) [#permalink]
09 Aug 2010, 11:11
You get multiple results for both statements, ie you cannot find a single (sifficient) value for x.
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Re: sqrts and primes (m06q31) [#permalink]
09 Aug 2010, 11:17
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
(a) implies N+1=squareroot(x+1)
implies x=(N+1)^2-1
We then calculate x for each prime number N which also verify 0<x<53 (see below)
N x
2 3
3 15
5 35
7 63
Three solutions, 3, 15 and 35 out of which 15 and 35. Hence 1 and 2 togethr are insufficient
let N+1=rootsquare (x+1) --> from 2 N is a prime number and x can be expressed as:
x=(N+1)^2-1
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Re: sqrts and primes (m06q31) [#permalink]
09 Aug 2010, 12:06
If 0<x<53, what is the value of integer ?
1. x is divisible by at least 2 prime numbers greater than 2
2. sq rt(x+1) -1 is prime
Prime numbers we can think of is 2, 3,5,7.11,13,17
Thus,
From 1 x is not divisible by 2 Thus x can be divisible by 3,5,7,11,13,17
Thus combination can be,
3,5
3,7
3,11
3,13
3,17
5,7
Now From 2: sq rt(x+1) -1 is a Prime number
Thus 2 values 15(3,5)/35(5,7)
Thus 2 solutions (3,5) & (5.7) Thus both State are insuff.
E
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Re: sqrts and primes (m06q31) [#permalink]
10 Aug 2010, 12:28
+1 for E.
Not an Unique value is coming for x
Cheers!
Ravi
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Re: sqrts and primes (m06q31) [#permalink]
10 Aug 2011, 07:13
E. great work in the first post
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Re: sqrts and primes (m06q31) [#permalink]
10 Aug 2011, 10:51
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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: X / under radical / the whole statement / Prime? 3 / 4 / 1 / No 8 / 9 / 2 / Yes15 / 16 / 3 / Yes24 / 25 / 4 / No35 / 36 / 5 / Yes48 / 49 / 6 / NoHopefully, we can't get further since x<53. So, according to statement2, the only possible values for x are 35,8,15 b/c these are the only numbers that can make statement 2 a prime Now, let's get to statement 1. It can leave out 8 from the list above but still we can't decide b/w 15 and 35. So E is the answer
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Re: sqrts and primes (m06q31) [#permalink]
10 Aug 2011, 11:53
E
Multiple of 3, 5, 7, 11, 13, 17 are 15, 21, 33, 35, 39, 51
1-Insufficient
2-Insufficient
now both statements answer can 3- for 15 but 5 for 35.. multiple answers .. so insufficient
Answer E
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Re: sqrts and primes (m06q31) [#permalink]
10 Aug 2011, 17:14
E is the answer....good one
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Re: sqrts and primes (m06q31) [#permalink]
14 Aug 2011, 16:52
E Posted from my mobile device 
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Re: sqrts and primes (m06q31) [#permalink]
14 Aug 2012, 06:24
Solved it but took 3 mins
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Re: sqrts and primes (m06q31) [#permalink]
14 Aug 2012, 07:13
bmwhype2 wrote: If 0 \lt x \lt 53 , what is the value of integer x ? 1. x is divisible by at least 2 prime numbers greater than 2 2. \sqrt{x +1} - 1 is prime Source: GMAT Club Tests - hardest GMAT questions Please show your work especially for 2. thanks (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.
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Re: sqrts and primes (m06q31) [#permalink]
15 Aug 2012, 08:29
1) 3*5=15, 3*7=21, 7*5=35, 11*3=33.... INSUFFICIENT
2) (8+1)^(1/2)-1=2
(15+1)^(1/2)-1=3 .... INSUFFICIENT
1&2) X could be 15 or 35... INSUFFICIENT
Answer: E
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Re: sqrts and primes (m06q31) [#permalink]
29 Aug 2012, 07:44
Took me longer that I initially though and close to 2 mins. Followed the approach of plugging in numbers. (i). x can be 15,21,33,35 (ii).x can be 8,15,35 Thus both together do not held us to conclude an unique value of x. Hence I chose answer choice E.
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Re: sqrts and primes (m06q31)
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29 Aug 2012, 07:44
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