A number has exactly 15 composite factors. What can be the maximum number of prime factors of this number Select one: a. 3 b. 4 c. 5 d. 2
Answers
Given : A number has exactly 15 composite factors.
To Find : What can be the maximum number of prime factors of this number
Solution:
Let say Prime factors are 2
m & n
=>
Number = mᵃ.nᵇ
Total Factors = (a + 1)(b + 1) a, b ≥ 1
Composite factors = 15
Prime factors = 2
Neither composite nor prime factor = 1
Total factor = 18
=> (a + 1)(b + 1) = 18
possible ordered pairs of a , b (1 , 8) , ( 2 , 5)
Let say Prime factors are 3
m , n & p
=>
Number = mᵃ.nᵇp^c
Total Factors = (a + 1)(b + 1)(c+1) a, b,c ≥ 1
Composite factors = 15
Prime factors = 3
Neither composite nor prime factor = 1
Total factor = 19
=> (a + 1)(b + 1)(c+1) = 19
Not possible
Let say Prime factors are 4
m , n , p & q
=>
Number = mᵃ.nᵇp^c .q^d
Total Factors = (a + 1)(b + 1)(c+1)(d+1) a, b,c,d ≥ 1
Composite factors = 15
Prime factors = 4
Neither composite nor prime factor = 1
Total factor = 20
=> (a + 1)(b + 1)(c+1)(d+1) = 20
20 = 2 * 2 * 5 or 4 * 5
Hence not possible
prime factors are 5
(a + 1)(b + 1)(c+1)(d+1)(e + 1) = 21
21 = 3 * 7
Hence not possible
So maximum prime factors = 2
Example :
2⁵ * 3² = 288
Composite factors = 15
4 , 6 , 8 , 9 , 12 , 16 , 18 , 24 , 32 , 32 , 36 , 48 , 72 , 144 , 288
Prime factor = 2
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