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A number N^2 has 35 factors. How many factors can N have?
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20 Oct 2017, 05:07
20 Oct 2017, 05:07
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A number \(N^2\) has 35 factors. How many factors can \(N\) have?
A. 6 or 10 factors
B. 8 or 14 factors
C. 10 or 16 factors
D. 12 or 18 factors
E. 14 or 20 factors
A. 6 or 10 factors
B. 8 or 14 factors
C. 10 or 16 factors
D. 12 or 18 factors
E. 14 or 20 factors
A number N^2 has 35 factors. How many factors can N have?
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20 Oct 2017, 06:22
20 Oct 2017, 06:22
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Expert Reply
Mahmud6 wrote:
A number \(N^2\) has 35 factors. How many factors can \(N\) have?
A. 6 or 10 factors
B. 8 or 14 factors
C. 10 or 16 factors
D. 12 or 18 factors
E. 14 or 20 factors
A. 6 or 10 factors
B. 8 or 14 factors
C. 10 or 16 factors
D. 12 or 18 factors
E. 14 or 20 factors
Hi...
Factors of 35 are 1,5,7,35 .. 35 = 1*35= 5*7..
So we can look for two cases..
1) where N or \(N^2\) contains ONLY one type of prime factor, a, that is \(N=a^x\)
Number of factors =(power of factor +1)=35.......
In this case power of the prime factor will be TWO times that of N as
\(N^2=a^2x....... So 2x+1=35....x=17..\\
N=a^17\)
So factor of N =17+1=18
2) where N^2 contains TWO prime factors...a and b.
So \(N=a^x*b^y..........N^2=a^2x*b^2y..\)
Number of factors=\((2x+1)(2y+1)=35=5*7\).....
Only possibility is when 2x+1=5....x=2
And \(2y+1=7...y=3\)
So \(N=a^2*b^3\)..
Number of factors=(2+1)(3+1)=3*4=12
Ans 12 and 18
D
General Discussion
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A number N^2 has 35 factors. How many factors can N have?
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20 Oct 2017, 05:27
20 Oct 2017, 05:27
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Lets take an example 12. Factors of 12 are 2- 2time and 3- 1 time. Total no of factor - (2+1) * (1+1) =6.
For 12*12 - it has 2 - 4times and 3 - 2 timed. Total no of factor - 5*3=15
Case 1 :-
N=x.y
For N*N
Trying reverse approach. 35 = 5*7.
Factors are x - 4times and y- 6times.
FOR N - x -2 times and y- 3 times.
Total no of factor is (2+1)*(3+1)=12
Case 2:-
If N^2 is raised to the power of 34.
N^2 will have 34 factors.
And hence for N we have 17 factors.
Hence total no of factor as 17+1 = 18
Ans is D
For 12*12 - it has 2 - 4times and 3 - 2 timed. Total no of factor - 5*3=15
Case 1 :-
N=x.y
For N*N
Trying reverse approach. 35 = 5*7.
Factors are x - 4times and y- 6times.
FOR N - x -2 times and y- 3 times.
Total no of factor is (2+1)*(3+1)=12
Case 2:-
If N^2 is raised to the power of 34.
N^2 will have 34 factors.
And hence for N we have 17 factors.
Hence total no of factor as 17+1 = 18
Ans is D
