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29 Aug 2023, 08:00
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D
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Difficulty:
35%
(medium)
Question Stats:
65% (01:15) correct
35%
(01:40)
wrong
based on 186
sessions
History
Date
Time
Result
Not Attempted Yet
If n is the product of the 5 different prime numbers, how many factors n have except 1 and n?
A. 20
B. 22
C. 24
D. 30
E. 32
A. 20
B. 22
C. 24
D. 30
E. 32
29 Aug 2023, 08:21
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Bunuel
If the prime factorization of N = (p^a)(q^b)(r^c) . . . (where p, q, r, etc are different prime numbers), then N has a total of (a+1)(b+1)(c+1)(etc) positive divisors.
Example: 14,000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14,000 = (4+1)(3+1)(1+1) =(5)(4)(2) = 40
For this question, n = (p^1)(q^1)(r^1)(s^1)(t^1) where p, q, r, s and t are the five prime numbers.
So, applying the formula, the number of positive divisors of n = (1+1)(1+1)(1+1)(1+1)(1+1) =(2)(2)(2)(2)(2) = 32
Since we are NOT including two factors (1 and n), the total number of divisors = 32 - 2 = 30
Answer: D
29 Aug 2023, 09:46
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Number(n) = a*b*c*d*e (1 is not considered)
We have to understand the pattern of factors- Factors of would be a, b, c, d, e, ab, ac, ad ..., abc, abd, abe, ...., abcd, abce, ..., abcde. so on.
Basically, 5C2 + 5C2 + 5C3 + 5C4. (5C5 corresponds to n)
5 + 10 + 10 + 5 = 30
Pray this user scores a 730, thanks.
We have to understand the pattern of factors- Factors of would be a, b, c, d, e, ab, ac, ad ..., abc, abd, abe, ...., abcd, abce, ..., abcde. so on.
Basically, 5C2 + 5C2 + 5C3 + 5C4. (5C5 corresponds to n)
5 + 10 + 10 + 5 = 30
Pray this user scores a 730, thanks.
