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12 Oct 2021, 05:45
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
68% (02:29) correct
32%
(02:28)
wrong
based on 175
sessions
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If X and Y are positive integers such that the least common multiple of \(X^2\) and \(Y^2\) is 3600 and the greatest common factor of \(X^2\) and \(Y^2\) is 9, how many distinct factors does 21*X*Y have?
A. 4
B. 6
C. 18
D. 48
E. 180
A. 4
B. 6
C. 18
D. 48
E. 180
12 Oct 2021, 07:09
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LCM x HCF = X^2 x Y^2
=> X^2 x Y^2 = 3600 x 9
=> (XY)^2 = 3600 x 9
=> XY = (3600 x 9)^(1/2) = 180
Hence 21 * X * Y = 180 * 21
=> 21 * X * Y = 3^3 * 2^2 * 5 * 7
Hence factors would be 4 * 3 * 2 * 2 = 48
=> X^2 x Y^2 = 3600 x 9
=> (XY)^2 = 3600 x 9
=> XY = (3600 x 9)^(1/2) = 180
Hence 21 * X * Y = 180 * 21
=> 21 * X * Y = 3^3 * 2^2 * 5 * 7
Hence factors would be 4 * 3 * 2 * 2 = 48
23 Oct 2021, 05:38
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Given that LCM of\( X^2\) and \(Y^2 \)is 3600 . So, LCM of X and Y must be 60
and given also that HCF of \( X^2\) and \(Y^2 \) is 9, so HCF of X and Y must be 3
As we know that Product of \(X*Y=LCM(X,Y) * HCF(X,Y)\)
=> \(X*Y = 60*3= 2^2*3^2*5^1\)
The question asks - how many distinct factors does 21*X*Y have?
\(21*X*Y = 7*3*X*Y = 2^2*3^3*5^1*7^1\)
Hence distinct factors would be \((2+1)*(3+1)*(1+1)*(1+1)\) \(=>\) \( 3 *4* 2 * 2 = 48\)
So Answer is Option D.
and given also that HCF of \( X^2\) and \(Y^2 \) is 9, so HCF of X and Y must be 3
As we know that Product of \(X*Y=LCM(X,Y) * HCF(X,Y)\)
=> \(X*Y = 60*3= 2^2*3^2*5^1\)
The question asks - how many distinct factors does 21*X*Y have?
\(21*X*Y = 7*3*X*Y = 2^2*3^3*5^1*7^1\)
Hence distinct factors would be \((2+1)*(3+1)*(1+1)*(1+1)\) \(=>\) \( 3 *4* 2 * 2 = 48\)
So Answer is Option D.
