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14 Apr 2020, 03:53
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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:
65%
(hard)
Question Stats:
50% (01:39) correct
50%
(02:11)
wrong
based on 209
sessions
History
Date
Time
Result
Not Attempted Yet
How many positive factors of \(2^5*3^6*5^2\) are perfect squares?
A. 8
B. 12
C. 18
D. 24
E. 36
Are You Up For the Challenge: 700 Level Questions
A. 8
B. 12
C. 18
D. 24
E. 36
Are You Up For the Challenge: 700 Level Questions
14 Apr 2020, 04:12
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Bunuel
How many positive factors of \(2^5*3^6*5^2\) are perfect squares?
A. 8
B. 12
C. 18
D. 24
E. 36
A. 8
B. 12
C. 18
D. 24
E. 36
CONCEPT: Perfect squares have Even exponents of all distinct prime factors of the Number
\(2^5*3^6*5^2 = (2^2)^2*(3^2)^3*(5^2)^1*2\)
\(2^2 = a\)
\(3^2 = b\)
\(5^2 = c\)
The factors will be perfect squares if we find the factors of \((a)^2*(b)^3*(c)^1\)
If \(N = a^p*b^q*c^r...\)
where a, b, c... are distinct primes
Number of factors of \(N = (p+1)*(q+1)*(r+1)*...\)
where a, b, c... are distinct primes
Number of factors of \(N = (p+1)*(q+1)*(r+1)*...\)
i.e. Factors of \((a)^2*(b)^3*(c)^1\) will be \((2+1)*(3+1)*(1+1) = 24\)
Answer: Option D
To know the derivation of Property of ways of finding factors of a number is available in th video here
General Discussion
14 Apr 2020, 04:02
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Make pairs of the squares!
Showing (2^4)*(3^6)*(5^2) as (2^2)*(3^3)*(5^1) (i.e. (2^a)*(3^b)*(5^c))
No. Of factors = (a+1)*(b+1)*(c+1) = (2+1)*(3+1)*(1+1) = 24
Showing (2^4)*(3^6)*(5^2) as (2^2)*(3^3)*(5^1) (i.e. (2^a)*(3^b)*(5^c))
No. Of factors = (a+1)*(b+1)*(c+1) = (2+1)*(3+1)*(1+1) = 24
