different positive divisors

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different positive divisors

by eshwarjayanth » Tue May 10, 2011 3:29 am
X=12Y, where Y is a prime number greater than 3. How many different positive divisors does X^2 having including X.

7
12
25
35
45

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by manpsingh87 » Tue May 10, 2011 3:40 am
eshwarjayanth wrote:X=12Y, where Y is a prime number greater than 3. How many different positive divisors does X^2 having including X.

7
12
25
35
45
X=2^2*3*y;
x^2= 2^4*3^2*y^2;
as y is prime therefore total no. of positive divisors of x is (4+1)*(2+1)*(2+!1)= 5*3*3=45;
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by bubbliiiiiiii » Tue May 10, 2011 4:06 am
therefore total no. of positive divisors of x is (4+1)*(2+1)*(2+!1)= 5*3*3=45;
How did you conclude this?
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by djiddish98 » Tue May 10, 2011 4:20 am
I believe the addition of the 1 for each term accounts for the fact that you can raise a number to 0.

So your options for the first prime (2) can be raised between 0-4, or 5 total options.

the second prime (3) can be raised between 0-2.

the third prime (y) can be raised between 0-2.

5 options * 3 options * 3 options.

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by GMATGuruNY » Tue May 10, 2011 4:57 am
eshwarjayanth wrote:X=12Y, where Y is a prime number greater than 3. How many different positive divisors does X^2 having including X.

7
12
25
35
45
To count the number of positive factors of an integer:

1) Prime-factorize the integer
2) Add 1 to each exponent
3) Multiply


For example:
72 = 2³ * 3².
Adding 1 to each exponent and multiplying, we get (3+1)(2+1) = 12 factors.

Here's the reasoning. To determine how many factors can be formed from 72 = 2³ * 3², we need to determine the number of choices we have of each prime factor:

For 2, we can use 2�, 2¹, 2², or 2³, giving us 4 choices.
For 3, we can use 3�, 3¹, or 3², giving us 3 choices.

Multiplying, we get 4*3 = 12 possible factors.

To verify, here are the 12 positive factors of 72:
1 72
2 36
3 24
4 18
6 12
8 9

Onto the problem above:

Plug in y=5.
Then x = 12 * 5 = 2² * 3 * 5
x² = (2² * 3 * 5)² = 2� * 3² * 5².

Adding 1 to each exponent and multiplying, we get:
Number of factors = (4+1)(2+1)(2+1) = 5*3*3 = 45.

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by eshwarjayanth » Tue May 10, 2011 6:18 am
Thanks all..