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14 Apr 2020, 03:55
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
74% (00:53) correct
26%
(01:43)
wrong
based on 231
sessions
History
Date
Time
Result
Not Attempted Yet
How many positive factors of \(2^4*5^3*7^4\) are odd numbers?
A. 100
B. 99
C. 24
D. 22
E. 20
Are You Up For the Challenge: 700 Level Questions
A. 100
B. 99
C. 24
D. 22
E. 20
Are You Up For the Challenge: 700 Level Questions
14 Apr 2020, 04:05
Kudos
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Bunuel
Concept:For ODD factors we need to exclude all powers of 2
i.e. Odd factors of \(2^4*5^3*7^4\) will be formed by only \(5^3*7^4\)
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 \(N = (p+1)*(q+1)*(r+1)*...\)
Total factors of \(5^3*7^4\) \(= (3+1)*(4+1) = 20\)
Answer: Option E
To know the derivation of Property of ways of finding factors of a number is available in th video here
14 Apr 2020, 04:40
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Bunuel
Solution:
- • To solve this question and find the number of odd factors, we need to ignore the prime number 2 and it's power
• So, we are left with 5^3 * 7^4.
• Now, the of odd factors = (Power of 5 +1)* (power of 7+1)=\((3+1)(4+1)=4*5=20\).
