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M06-31

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:28
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Difficulty:

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Question Stats:

60% (01:24) correct 40% (01:27) wrong based on 121 sessions

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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

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Kudos [?]: 135518 [1], given: 12697

Math Expert
Joined: 02 Sep 2009
Posts: 42583

Kudos [?]: 135518 [1], given: 12697

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15 Sep 2014, 23:28
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Official Solution:

Statement (1) by itself is insufficient.

Statement (2) by itself is insufficient. Run squares of 2, 3, 4, 5, 6, and 7. In reality we need only squares of 2, 4, 6 as $$x -1$$ is prime and thus, the square root result has to be even to make $$x - 1$$ odd. If $$x = 15$$, $$\sqrt{16} = 4$$ and $$4 - 1 = 3$$. Also, if $$x = 35$$, $$\sqrt{36} = 6$$, and $$6 - 1=5$$. There are several values that work for $$x$$.

Statements (1) and (2) combined are insufficient. If we take both statements together, $$x$$ can be 35 or 15. Both are divisible by several primes greater than 2.

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Joined: 16 Oct 2016
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13 Mar 2017, 00:26
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient.

Statement (2) by itself is insufficient. Run squares of 2, 3, 4, 5, 6, and 7. In reality we need only squares of 2, 4, 6 as $$x -1$$ is prime and thus, the square root result has to be even to make $$x - 1$$ odd. If $$x = 15$$, $$\sqrt{16} = 4$$ and $$4 - 1 = 3$$. Also, if $$x = 35$$, $$\sqrt{36} = 6$$, and $$6 - 1=5$$. There are several values that work for $$x$$.

Statements (1) and (2) combined are insufficient. If we take both statements together, $$x$$ can be 35 or 15. Both are divisible by several primes greater than 2.

Can you please explain why statement 2 is insufficient? Also the question does not say anything about $$x -1$$ being a prime number.

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13 Mar 2017, 00:29
kaustav.becs@gmail.com wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient.

Statement (2) by itself is insufficient. Run squares of 2, 3, 4, 5, 6, and 7. In reality we need only squares of 2, 4, 6 as $$x -1$$ is prime and thus, the square root result has to be even to make $$x - 1$$ odd. If $$x = 15$$, $$\sqrt{16} = 4$$ and $$4 - 1 = 3$$. Also, if $$x = 35$$, $$\sqrt{36} = 6$$, and $$6 - 1=5$$. There are several values that work for $$x$$.

Statements (1) and (2) combined are insufficient. If we take both statements together, $$x$$ can be 35 or 15. Both are divisible by several primes greater than 2.

Can you please explain why statement 2 is insufficient? Also the question does not say anything about $$x -1$$ being a prime number.

(2) says that $$\sqrt{x +1} - 1$$ is prime. This statement is not sufficient because it holds true for different values of x. For example, x can be 35 or 15.
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14 Mar 2017, 10:25
kaustavbecs wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient.

Statement (2) by itself is insufficient. Run squares of 2, 3, 4, 5, 6, and 7. In reality we need only squares of 2, 4, 6 as $$x -1$$ is prime and thus, the square root result has to be even to make $$x - 1$$ odd. If $$x = 15$$, $$\sqrt{16} = 4$$ and $$4 - 1 = 3$$. Also, if $$x = 35$$, $$\sqrt{36} = 6$$, and $$6 - 1=5$$. There are several values that work for $$x$$.

Statements (1) and (2) combined are insufficient. If we take both statements together, $$x$$ can be 35 or 15. Both are divisible by several primes greater than 2.

Can you please explain why statement 2 is insufficient? Also the question does not say anything about $$x -1$$ being a prime number.

We can simplified Statement 2 as follows:

$$\sqrt{x +1} - 1$$ = prime

$$\sqrt{x +1}$$ = prime + 1 ...square both sides

x + 1= (Prime+1)^2

x + 1= Prime^2 + 2 Prime + 1

x = Prime^2 + 2 Prime

Choose prime as follows (easy task)

Let Prime=2 ........x = 8

Let Prime= 3........x=15

Let Prime= 5........x=35

You can't put prime= 7 because this will violate the question that 0<x<53

When you combine 1 & 2

You will have x= 15 that has two prime numbers 3 & 5
or
You will have x= 35 that has two prime numbers 5 & 7

I hope it help

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10 Nov 2017, 18:38
for evaluating Stmt 2, I wonder why you skipped 8?
sqrt(8+1) - 1 = 3-1 = 2 which is a prime number. So the possible x values are 8, 15, 35.

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11 Nov 2017, 00:18
sevenplusplus wrote:
for evaluating Stmt 2, I wonder why you skipped 8?
sqrt(8+1) - 1 = 3-1 = 2 which is a prime number. So the possible x values are 8, 15, 35.

____________________
Yes, 8 works too but there are more than three values possible and it does not matter which two you choose to get insufficiency.
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Re: M06-31   [#permalink] 11 Nov 2017, 00:18
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