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Here is what i did in this Question.
We are asked about the number of factors of a Positive integer N.

Statement 1->
The only way 27*N^3 =3^3*N^3 will have 16 factors is if N is a prime number.
Hence N must be a prime number => 2 factors only.
Hence Sufficient.

Statement 2->
Since N is an integer => N^3 must be a perfect Cube.
The only perfect cube between 90 and 200 is 125.
Hence N^3 must be 125 => N must be 5 => Two factors only.
Hence sufficient.

Hence D.
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If \(N\) is a positive integer,what is the value of \(N\)?
(1) \(N\) is divisible by 31.
(2) \(\frac{N}{100}\) is a prime number.



Source->Self-Made
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If a positive integer \(n\) is given by \(n=26*p\) where \(p\) is a prime number greater than 2,how many even divisors does n have ?

A)4
B)3
C)2
D)1
E)cannot be determined.




Source-> Self-Made

n=26p
=2*13*p
where p is odd(prime >2 is odd)
thus no. of even factors are
2 , 2*13 , 2*p , 2*13*p
=4

Ans A
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stonecold
If a positive integer \(n\) is given by \(n=26*p\) where \(p\) is a prime number greater than 2,how many even divisors does n have ?

A)4
B)3
C)2
D)1
E)cannot be determined.




Source-> Self-Made

n=26p
=2*13*p
where p is odd(prime >2 is odd)
thus no. of even factors are
2 , 2*13 , 2*p , 2*13*p
=4

Ans A

Counter Question->
What if p=13?

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stonecold
If a positive integer \(n\) is given by \(n=26*p\) where \(p\) is a prime number greater than 2,how many even divisors does n have ?

A)4
B)3
C)2
D)1
E)cannot be determined.




Source-> Self-Made

n=26p
=2*13*p
where p is odd(prime >2 is odd)
thus no. of even factors are
2 , 2*13 , 2*p , 2*13*p
=4

Ans A

Counter Question->
What if p=13?


sure
p may be 13
then this is debatable one
on my side it will better be a DS question
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rohit8865


Counter Question->
What if p=13?


sure
p may be 13
then this is debatable one
on my side it will better be a DS question

We may not agree on that. :)
Since two values of the number of factors are possible => The answer cannot be determined.
Hence E.
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stonecold


sure
p may be 13
then this is debatable one
on my side it will better be a DS question
[/quote]

We may not agree on that. :)
Since two values of the number of factors are possible => The answer cannot be determined.
Hence E.
[/quote]

well i have not considered the last option
Ans will be E
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A number is said to be prime saturated if the product of all the different positive prime factors of n is less than the square root of n.
If p is a prime saturated,what is the value of p?

A)p is a perfect cube.
B)2≤p≤15



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If \(N\) is a positive integer,what is the value of \(N\)?
(1) \(N\) is divisible by 31.
(2) \(\frac{N}{100}\) is a prime number.



Source->Self-Made

(1) n could be 31 or 62
not suff

(2) n can be 200 or 300
not suff

combining n=31p from (1)
& 31p/100 is prime From (2)
thus n must be 3100

Ans C
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St1: x = y = z = 3 --> x * y * z is not divisible by 3*7
x = 3, y = z = 21 --> x * y * z is divisible by 3*7
Not Sufficient.

St2: LCM(x, y, z) = 63
x * y * z is a multiple of 63 and hence divisible by 21
Sufficient.

Answer: B
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A number is said to be prime saturated if the product of all the different positive prime factors of n is less than the square root of n.
If p is a prime saturated,what is the value of p?

A)p is a perfect cube.
B)2≤p≤15



Source-> Self-Made


A) \(p\) is a perfect cube. Then \(p\) could be \(3^3=27\) or \(2^3=8\). Insufficient.

B) \(2 \leq p \leq 15\)

First, \(p\) can't be a prime. Out 2, 3, 5, 7, 11, 13, left 4, 6, 8, 9, 10, 12, 14, 15
Second, \(p\) can't be in prime factorization of \(a \times b\) where \(a\) and \(b\) are prime. Out 6, 10, 14, 15. Left 4, 8, 9, 12.
Third, \(p\) can't be in form of \(a^2\) where \(a\) is prime. Out 4, 9. Left 8, 12.

8 is a cube, so 8 is prime saturated.
\(12=2^2 \times 3\). We have \(2 \times 3 = 6 > \sqrt{12}=2\sqrt{3}\) so 12 is not prime saturated.

Hence, \(p=8\). Sufficient.

The answer is B.
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If \(a = b^3*c^2*d\),how many positive factors does a have?

(1)\(b\), \(c\) and \(d\) are prime numbers.
(2)\(b\),\(c\),\(d\) are distinct integers.



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If \(a = b^3*c^2*d\),how many positive factors does a have?

(1)\(b\), \(c\) and \(d\) are prime numbers.
(2)\(b\),\(c\),\(d\) are distinct integers.



Source->Self-Made

Clearly, the answer is A
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If \(x\) is an integer, is \(x^3\) divisible by 7 ?
(1) \(x^{12}\) is divisible by 7.
(2) \(x^4\) is divisible by 7.
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How many positive even divisors does 96 have ?

A. 12
B. 10
C. 8
D. 6
E. Cannot be determined

96= 2^5*3
thus no. of even factors = 5*2=10

Ans B
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Mock Test 1(600-700) Updated.

stonecold-s-mock-test-217160.html#p1676182
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Mock Test 2 (700+) Updated.


stonecold-s-mock-test-217160.html#p1676182
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