avinashrao9 wrote:
mau5 wrote:
abhishekkhosla wrote:
How many natural number less than 200 will have 12 factors?
a)2
b)4
c)6
d)8
e)none of these
The no of factors is 12 : The given no could be of the form : \(a*b*c^2\) or \(a^5*b\) or \(a^3*b^2 or a^{11}\), where a,b,c are primes
Now, the minimum integer which is of the form \(a*b*c^2=3*5*2^2 = 60\)
Similarly, you could fix c at 2, and we could have\(3*7*2^2,3*11*2^2,3*13*2^2\) : that makes it 4 numbers which are all less than 200.
Again, fixing c=3, we have \(2*5*3^2,2*7*3^2,2*11*3^2\)= 3 integers.
Again, we could have c=5 and we have \(2*3*5^2 = 1\) integer. We anyways have 4+3+1 = 8 integers.
There will be more such nos for example \(2^5*3 , 2^3*3^2,\)etc.
Thus, the answer will be E.
Could you please explain how did you arrive at
\(a*b*c^2\) or \(a^5*b\) or \(a^3*b^2 or a^{11}\)
I am confused
From the GmatClub Book, Number theory(
math-number-theory-88376.html)
Finding the Number of Factors of an Integer
First make prime factorization of an integer \(n=a^p*b^q*c^r\), where a, b, and c are prime factors of n and p, q, and r are their powers.
The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450:\(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is (1+1)*(2+1)*(2+1)=2*3*3=18 factors.
Back to the problem:As we are given that the no of factors is 12, we know that the given number will have a prime factorization where in (p+1)(q+1)(r+1)... = 12.
12 can be written as 12*1 = (p+1), p =11
6*2 = (p+1)*(q+1), p=5,q=1
4*3 = (p+1)(q+1), p = 3,q = 2
2*2*3 = (p+1)(q+1)(r+1) , p=q=1,r=2
Hope this helps. Please ask if anything remains unclear.
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