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RenB
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Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
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09 May 2023, 19:35
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Was unaware about this shortcut- Sharing FYR
Take any number “N” and convert it into product of prime numbers (Prime factorization) i.e
\(N = A^p x B^q x C^r\)
Here A, B , C are prime numbers and p,q,and r were respective powers of that prime numbers.
Total numbers of factors for N = (p + 1)(q +1)(r +1)
Say A is an even prime number (2) and B and C are odd prime numbers.
The total number of even factors= (p+1)
The total number of odd factors= (q+1)(r+1)
Or Total number of odd factors= Total factors- Even factors= (p + 1)(q +1)(r +1)- (p+1)
E.g.= Total, odd and even factors of 4500
First write the number 4500 into prime factorization
4500 = 45 x 100 = 9 x 5 x 10 x 10 = 3 x 3 x 5 x 5 x 2 x 5 x 2
4500 = 22 x 32 x 53 Here consider A = 2 , B = 3 , C = 5 , p= 2 , q = 2 and r = 3
Here identifying that odd number are 3 and 5
Numbers of odd factors of number 4500 = (q + 1 ) (r + 1) = 3 x 4 = 12
Total number of factors = (p + 1)(q +1)(r +1) = 3 x 3 x 4 =36
Numbers of even factors of number = Total number of factors – Numbers of odd factors = 36 – 12 = 24
Source- https://www.allmathtricks.com/factors-number/
Take any number “N” and convert it into product of prime numbers (Prime factorization) i.e
\(N = A^p x B^q x C^r\)
Here A, B , C are prime numbers and p,q,and r were respective powers of that prime numbers.
Total numbers of factors for N = (p + 1)(q +1)(r +1)
Say A is an even prime number (2) and B and C are odd prime numbers.
The total number of even factors= (p+1)
The total number of odd factors= (q+1)(r+1)
Or Total number of odd factors= Total factors- Even factors= (p + 1)(q +1)(r +1)- (p+1)
E.g.= Total, odd and even factors of 4500
First write the number 4500 into prime factorization
4500 = 45 x 100 = 9 x 5 x 10 x 10 = 3 x 3 x 5 x 5 x 2 x 5 x 2
4500 = 22 x 32 x 53 Here consider A = 2 , B = 3 , C = 5 , p= 2 , q = 2 and r = 3
Here identifying that odd number are 3 and 5
Numbers of odd factors of number 4500 = (q + 1 ) (r + 1) = 3 x 4 = 12
Total number of factors = (p + 1)(q +1)(r +1) = 3 x 3 x 4 =36
Numbers of even factors of number = Total number of factors – Numbers of odd factors = 36 – 12 = 24
Source- https://www.allmathtricks.com/factors-number/
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