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# X is prime and y is positive integer, How many different

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X is prime and y is positive integer, How many different [#permalink]

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11 Dec 2005, 06:36
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X is prime and y is positive integer, How many different factors of (2^3)*(x^y) are there?

1) x=5
2) y=3
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Re: DS - Factors [#permalink]

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11 Dec 2005, 08:56
B. since y is 3, its enough to find the factors of a given integer.

my 2kth post.
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11 Dec 2005, 09:22
CONGRATULATIONS himalaya..you are veteran now

for this question, i will go with C.

2^3 * x^y

S2 is insufficient:
if x = 2, y = 3
total factors: 7

x=5, y = 3
total factors: 16
insufficient

we need both.
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11 Dec 2005, 09:56
duttsit wrote:
CONGRATULATIONS himalaya..you are veteran now

for this question, i will go with C.

2^3 * x^y

S2 is insufficient:
if x = 2, y = 3
total factors: 7

x=5, y = 3
total factors: 16
insufficient
we need both.

yes, i overlooked that one.............

weldone, your math is near to perfect, dutt!!!!!!!!!!!!
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11 Dec 2005, 17:33
Congrats HIMALAYA! :

Yup, answer is C for this one

(2^3)*(2*3) = 2^6 ==> 7 factors
(2^3)*(5^3) = 4*4 ==> 16 factors
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Re: DS - Factors [#permalink]

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12 Dec 2005, 02:49
gamjatang wrote:
X is prime and y is positive integer, How many different factors of (2^3)*(x^y) are there?

1) x=5
2) y=3

How many different factors of (2^3)*(x^y) are there?
Always just two DIFFERENT: 2 and x
But x could be 2.
So
1) Is suff , just 2 factors
2) Is insuff, because x could be 2 or not, so it could be 1 or 2 different factors.
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Re: DS - Factors [#permalink]

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12 Dec 2005, 10:33
HIMALAYA wrote:
B. since y is 3, its enough to find the factors of a given integer.

my 2kth post.

The Answer is (C).

Congratulations on your 2000th post, HIMALAYA!!!
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13 Dec 2005, 14:51
C since it isn't give that x is a distinct prime factor other than 2.
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13 Dec 2005, 15:10
It's B

Number of factors =(3+1)(3+1)=12

For any number that is written as the product of primes a^p*b^r*c^q,
where a,b,c are primes, and p,r,q are their powers, the number of factors
=(p+1)(r+1)(q+1).
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14 Dec 2005, 08:59
B was the trap answer. C by brute force.
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14 Dec 2005, 10:22
C for me...cause (2) tells us the value of y...but we dont know if X =2 or some other prime...

(1) well we dont know the value of y, so we dont know how many factors...

1+2...sufficient..
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12 Jan 2006, 19:15
sorry to bring this old question up once again, but can someone explain how we know how many factors there are in , for example, 2^6 ?
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12 Jan 2006, 23:44
Well this is how we calculate..
If a number is raised to power of a prime number then the factor is n+1
e.g
2^3 means factors (3+1=4)
Lets see 2^3=8 (1,2,4,8)

Enother eg 2^3 * 5^4
Factors(3+1)*(4+1)=4*5=20
the catch here is tht the numbers should be prime...In our case both 2 and 5 are prime.
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13 Jan 2006, 02:33
One more thing that I would like to add is that the number of distinct product pairs can be given by 1/2 times the number of distinct factors ( except for numbers which are prefect squares).

andy_gr8 wrote:
Well this is how we calculate..
If a number is raised to power of a prime number then the factor is n+1
e.g
2^3 means factors (3+1=4)
Lets see 2^3=8 (1,2,4,8)

Enother eg 2^3 * 5^4
Factors(3+1)*(4+1)=4*5=20
the catch here is tht the numbers should be prime...In our case both 2 and 5 are prime.
[#permalink] 13 Jan 2006, 02:33
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