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13 Apr 2020, 07:57
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C
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Difficulty:
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Question Stats:
36% (02:31) correct
64%
(02:37)
wrong
based on 109
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How many un-ordered pairs of positive integers (x, y) exist such that the product of x, y and the greatest common factor of (x, y) is 1080?
A. 7
B. 8
C. 9
D. 10
E. 11
Are You Up For the Challenge: 700 Level Questions
A. 7
B. 8
C. 9
D. 10
E. 11
Are You Up For the Challenge: 700 Level Questions
13 Apr 2020, 08:44
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Bunuel
x*y* GCD(x,y) = 1080
But
PROPERTY: GCD (x,y) * LCM(x,y) = x*y
Also, x = GCD*a and y = GCD*b where a and b have no common factor other than 1
Also, x = GCD*a and y = GCD*b where a and b have no common factor other than 1
i.e. GCD*a * GCD*b * GCD = 1080 = 12*9*10 = 2^3*3^3^*5
i.e.\( GCD^3*a*b = 2^3*3^3*5\)
i.e. GCD = 1 or 2 or 3 or 6
i.e. a*b can be 1080, 135, 40 or 5 respectively
a number can be written as a product of two co-prime numbers in \(2^{n-1}\) ways where n is th number of Distinct prime factors of the Number
1080 (3 prime factors 2, 3, and 5) can be written as product of two coprime numbers in \(2^{3-1} = 4\) ways
135 (2 prime factors 3, and 5) can be written as product of two coprime numbers in \(2^{2-1} = 2\) ways
40 (2 prime factors 2, and 5) can be written as product of two coprime numbers in \(2^{2-1} = 2\) ways
5 (1 prime factors 5) can be written as product of two coprime numbers in \(2^{1-1} = 1\) ways
Total ways = 4+2+2+1 = 9
Answer: Option C
General Discussion
13 Apr 2020, 08:51
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GCF(x,y) = k
x=ka
y=kb, where a and b are coprimes
ka*kb*k = 1080
k^3 * a *b = 1080
k^3 * a *b = 2^3*3^3*5
k= 1; total ordered pairs are 2*(2^2)= 8
k= 2 or 3; total ordered pairs are 2*(2^1) = 4
k = 2*3; total ordered pairs are 2*(2^0) =2
Total ordered pairs are 8+4+4+2=18
Bunuel i guess there is a typo. It should be unordered pairs imo.
x=ka
y=kb, where a and b are coprimes
ka*kb*k = 1080
k^3 * a *b = 1080
k^3 * a *b = 2^3*3^3*5
k= 1; total ordered pairs are 2*(2^2)= 8
k= 2 or 3; total ordered pairs are 2*(2^1) = 4
k = 2*3; total ordered pairs are 2*(2^0) =2
Total ordered pairs are 8+4+4+2=18
Bunuel i guess there is a typo. It should be unordered pairs imo.
Bunuel
