If x is an integer and 0 < x < 67, what is the value of x? : Data Sufficiency (DS)

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15 Nov 2019, 04:13

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If x is an integer and 0 < x < 67, what is the value of x?

(1) x is divisible by at least two prime numbers greater than 2.
(2) \(\sqrt{x+1}−1\) is prime.

(1) x is divisible by at least two prime numbers greater than 2. insufic

\(x=(15,35,63)=(5*3,5*7,7*3^2)…\)

(2) \(\sqrt{x+1}−1\) is prime. insufic

\(\sqrt{x+1}−1=p…x+1=(p+1)^2=perf.square\)
\(0<(p+1)^2-1<67…1<(p+1)^2<68…1<p+1≤8…0<p≤7\)
\(p=(2,3,5,7)…(p+1)^2=(9,15,36,64)\)
\(x+1=(9,15,36,64)…x=(8,14,35,63)\)

(1&2) insufic

\(x=(35,63)=(5*7,7*3^2)\)

Ans (E)

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15 Nov 2019, 07:52

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If x is an integer and 0 < x < 67, what is the value of x?

(1) x is divisible by at least two prime numbers greater than 2.
x div. by 3 and 5. x = 15 or 45 ..
OR
x div. by 5 and 7. x = 35

INSUFFICIENT.

(2) \(\sqrt{x+1}−1\) is prime.
\(\sqrt{x+1}−1 = p\)
\(\sqrt{x+1} = p + 1\)
x + 1 = \(p^2 + 1 + 2p\)
x = p * (p + 2)
So, x = 2 * (2 + 2) = 8 OR x = 3 * (3 + 2) = 15 OR x = 5 * (5 + 2) = 35

INSUFFICIENT.

Together 1 and 2
Two possibilities still exists, x = 15 or x = 35. Hence

INSUFFICIENT.

IMO Answer E.

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15 Nov 2019, 11:00

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(1) x is divisible by at least two prime numbers greater than 2.
There are many numbers b/w 0 and 67.... INSUFFICIENT

(2) √x+1−1 is prime.

√63+1 -1 = 7
√35+1 -1 = 5
√8+1 -1 =3

So insufficient

Combining both only 35 is having two prime factors greater than 2

OA:C

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15 Nov 2019, 12:21

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(1) x is divisible by at least two prime numbers greater than 2.

The prime numbers could be 3,5,7,11,13 etc . Therefore numbers could be 15, 21,35,65 Insufficient.

(2) \sqrt{x+1}-1 is prime.

Let the above expression equal to Y.

Y+1 = \sqrt{X+1}

(y+1)^2 = x+1

Since Y is any prime number 1,2,3,5,7, 11.. there could a multitude of values for x like 3, 8,15,35,63

Insufficient

Both statements together, we see X could be 15, 35,65 . Therefore insufficient

Answer is E

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15 Nov 2019, 15:03

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If x is an integer and 0 < x < 67, what is the value of x?

(1) x is divisible by at least two prime numbers greater than 2.
(2)\(\sqrt{x+1}\)−1 is prime.

(1) x can be 15, 30, 45, 60 (divisible by 3 and 5 ), or 35 (divisible by 5 and 7) or 33,66 (divisible by 3 and 11), 55 (divisible by 5 and 11), 21, 42 (divisible by 3 and 7). So, Not sufficient.

(2) x can be 15 (3 is the resulting integer) or 35 (5 is the resulting integer). Not sufficient

Combining both, x can be either 15 or 35. Not sufficient.
E is the answer.

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15 Nov 2019, 21:10

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We are given that x is divisible by at least two prime numbers greater than 2 and that 0<x<67.

Statement 1: x is divisible by at least two prime numbers greater than 2.
Possible values of x: 15, 21, 30, 33, 35, etc.
Statement 1 does not lead to a unique value of x, hence it is not sufficient.

Statement 2: √(x+1)-1 is prime
Possible values of x: 8, 15, 35, 63.
Statement 2 does not lead to a unique value of x, hence it is not sufficient.

1+2
The numbers that satisfy conditions 1 and 2 are: 15, 35, and 63.
Statements 1 and 2 when combined do not lead to a unique value of x. This means both statements even when combined are not sufficient.

The answer is, therefore, option E.

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16 Nov 2019, 05:17

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St 1 : clearly not sufficient.

St 2: Not sufficient.

Both : sufficient.

Considering that X is divisible by at least 2 primes >2, then we can test for values ..3,5,7 etc.

35 = 7×5 satisfies the condition.

√35+1-1 = 5. Prime no.

That's why (c)

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16 Nov 2019, 07:20

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IMO E
Statement 1 : x can be 15 / 21 / 35, both divisible by 3 & 5, 3 & 7 and 7 & 5 respectively. Not sufficient.
Statement 2 : sqrt 35+1 = 6 -1 = 5, which is a prime number or x = 15. sqrt 15+ 1 = 4 -1 = 3 . Not sufficient.
Combine both statements, we have x = 15 or 35.

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16 Nov 2019, 12:12

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Possible values for x are: x=3*5=15, x= 5*7=35, x=11*13=143, x=17*19=323, x=29*31=899, x=41*43=1763, x=59*61=3599.

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16 Nov 2019, 13:26

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If x is an integer and 0 < x < 67, what is the value of x?

(Statement1):
—> x could be 15(3*5 ), or 35( 5*7), or 63 (7*9)
Insufficient

(Statement2):
If x=8, then \((8+1)^{1/2} —1= 3–1=2\)(prime)

If x=15, then \((15+1)^{1/2}—1= 4–1=3\) (prime)

Insufficient

Taken together 1&2,
x could be 15(3*5)
—> \((15+1)^{1/2} —1= 4–1=3\) (prime)

x could be 35 (5*7)
—> \((35+1)^{1/2}—1= 6–1=5\) (prime)

Also, x could be 63

Insufficient

The answer is E.

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If x is an integer and 0 < x < 67, what is the value of x?

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