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sqrts and primes (m06q31)

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sqrts and primes (m06q31) [#permalink] New post 12 Nov 2007, 10:50
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Question Stats:

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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

[Reveal] Spoiler: OA
E

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Please show your work especially for 2. thanks
[Reveal] Spoiler: OA
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Re: sqrts and primes (m06q31) [#permalink] New post 12 Nov 2007, 13:54
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] New post 12 Nov 2007, 21:01
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] New post 09 Aug 2010, 07:15
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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] New post 09 Aug 2010, 08:35
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] New post 09 Aug 2010, 10:11
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] New post 09 Aug 2010, 10:17
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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)^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] New post 09 Aug 2010, 11:06
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] New post 10 Aug 2011, 09:51
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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 / Yes
15 / 16 / 3 / Yes
24 / 25 / 4 / No
35 / 36 / 5 / Yes
48 / 49 / 6 / No

Hopefully, 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] New post 10 Aug 2011, 10:53
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] New post 14 Aug 2012, 05:24
Solved it but took 3 mins
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Re: sqrts and primes (m06q31) [#permalink] New post 14 Aug 2012, 06:13
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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

[Reveal] Spoiler: OA
E

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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] New post 15 Aug 2012, 07:29
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] New post 29 Aug 2012, 06:44
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] New post 14 Aug 2013, 08:31
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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] New post 15 Aug 2013, 09:52
solved well in time..
answer E.
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Re: sqrts and primes (m06q31) [#permalink] New post 30 Jul 2014, 05:01
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)   [#permalink] 30 Jul 2014, 05:01
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