Find number of factors of this- N= 2^3,6^2,15^3,3^3,7^2
Answers
Answer:
given
2³,6²,15³,3³,7²
Step-by-step explanation:
Factors of a number N refers to all the numbers which divide N completely. These are also called divisors of a number.
Basic formula related to factors of a number:
These are certain basic formulas pertaining to factors of a number N, such that,
N= paqbrc
Where, p, q and r are prime factors of the number n.
a, b and c are non-negative powers/ exponents
Number of factors of N = (a+1)(b+1)(c+1)
Product of factors of N = N No. of factors/2
Sum of factors: ( p0+p1+...+pa) ( q0+ q1+....+qb) (r0+r1+...+rc)/ (pa-1)(qb-1)(rc-1)
Solved questions on Factors of a number:
Example 1: Consider the number 120. Find the following for n
Sum of factors
Number of factors
Product of factors
Solution: The prime factorization of 120 is 23x31x51. By applying the formulae,
Sum of factors = [(20+21+22+23)(30+31)(50+51)]/ [(2-1) (3-1)(5-1)] = 45
Number of factors = (3+1)(1+1)(1+1) = 16
Product of factors = 120(16/2) = 1208
Example 2: Find the following for the number 84 :-
Number of odd factors
Number of even factors
Solution: By the prime factorization of 84, 84= 22 × 31 × 71
Total number of factors = (2+1)(1+1)(1+1) = 12
Number of odd factors will be all possible combinations of powers of 3 and 5 (excluding any power of 2) . Hence number of odd factors = (1+1)(1+1) = 4
By manually checking, these factors are 1, 3, 7 and 21.
Number of even factors = total no. of factors - no. of even factors
= 12 - 4 = 8
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