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No of distinct pairs of multiples

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No of distinct pairs of multiples

by ronnie1985 » Thu Jun 21, 2012 4:52 am
For all positive integers f, fâ—Ž equals the distinct pairs of positive integer factors. For example, 16â—Ž =3, since there are three positive integer factor pairs in 16: 1 x 16, 2 x 8, and 4 x 4.

For which of the following values of f does fâ—Ž equal 4?

(A) 4
(B) 20
(C) 30
(D) 60
(E) 80

I want to know:What is the formula for calculating the number of distinct pairs of factors for a given number.

Thanks for help!!!
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by Anurag@Gurome » Thu Jun 21, 2012 5:05 am
ronnie1985 wrote:I want to know:What is the formula for calculating the number of distinct pairs of factors for a given number.
Number of distinct pairs of factors for a given integer = (Number of distinct factors of the integer)/2

And if the integer is a square of an integer then, number of distinct pairs of factors the integer = (Number of distinct factors of the integer + 1)/2

Now, number of factors of an integer whose prime factorization is of the form (p^a)*(q^b)*(r^c)... is (a + 1)(b + 1)(c + 1)...

Now coming to the question, let's check each of the options..
  • (A) 4 = (2^2) --> Number of factors = (2 + 1) = 3 --> Number of pairs = (3 + 1)/2 = 2
    (B) 20 = (2^2)*(5^1) --> Number of factors = (2 + 1)*(1 + 1) = 6 --> Number of pairs = 6/2 = 3
    (C) 30 = (2^1)*(3^1)*(5^1) --> Number of factors = (1 + 1)*(1 + 1)*(1 + 1) = 8 --> Number of pairs = 8/2 = 4 ---> ANSWER
    (D) 60 --> No need to calculate
    (E) 80 --> No need to calculate
The correct answer is C.
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