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sqrts and primes (m06q31) [#permalink]
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12 Nov 2007, 11:50
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This topic is locked. If you want to discuss this question please repost it in the respective forum. 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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Re: sqrts and primes (m06q31) [#permalink]
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12 Nov 2007, 14:54
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 (m06q31) [#permalink]
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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]
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09 Aug 2010, 08:15
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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]
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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]
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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]
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09 Aug 2010, 11:17
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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)^21
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)^21



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Re: sqrts and primes (m06q31) [#permalink]
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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]
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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]
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10 Aug 2011, 11:53
E Multiple of 3, 5, 7, 11, 13, 17 are 15, 21, 33, 35, 39, 51
1Insufficient 2Insufficient
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]
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14 Aug 2012, 06:24
Solved it but took 3 mins
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Re: sqrts and primes (m06q31) [#permalink]
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14 Aug 2012, 07:13
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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]
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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]
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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) [#permalink]
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14 Aug 2013, 09:31
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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.
Not Sufficient.



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Re: sqrts and primes (m06q31) [#permalink]
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15 Aug 2013, 10:52
solved well in time.. answer E.



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Re: sqrts and primes (m06q31) [#permalink]
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30 Jul 2014, 06:01
If 0 < x < 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
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.




Re: sqrts and primes (m06q31)
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