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STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.

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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 12 Jan 2017, 23:05
1
St1: k = 13t + 13
t = 0 --> k is prime
t = any other integer --> k is not prime
Not Sufficient

St2: k = 17! + 13 --> 13(Some number + 1) --> k is not prime
Sufficient

Answer: B
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 01:13
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 03:25
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 05:47
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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 05:55
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 07:43
stonecold wrote:
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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 07:47
rohit8865 wrote:
stonecold wrote:
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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 08:04
stonecold wrote:
rohit8865 wrote:
stonecold wrote:
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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 08:19
rohit8865 wrote:
stonecold wrote:
rohit8865 wrote:

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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 08:26
stonecold wrote:

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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 08:29
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
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 08:31
stonecold wrote:
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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 08:55
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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 09:17
1
stonecold wrote:
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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 13 Jan 2017, 10:33
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 14 Jan 2017, 05:07
stonecold wrote:
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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 14 Jan 2017, 09:18
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 14 Jan 2017, 11:11
stonecold wrote:
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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 15 Jan 2017, 14:17
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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New post 16 Jan 2017, 11:51
Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.   [#permalink] 16 Jan 2017, 11:51

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