Total number of factors of any number.

1.	Total number of factors of any number.
Answer» The number has to be resolved into prime factors. If the prime factorisation of the given number = 2^m × 3ⁿ × 5^p × 11^q , then the total number of factors = (m + 1) × (n + 1) × (p + 1) × (q + 1). For example, if N = 2⁴ × 5⁶× 7⁴ × 13¹ , then Number of factors of N = (4 + 1) × (6 + 1) × (4 + 1) × (1 + 1) = 5 × 7 × 5 × 2 = 350. 2⁴ = 16 \(\therefore\) factors of 16 are 1, 2, 4, 8, 16, i.e., 5 in number, i.e., (4 + 1) in numbers. Similarly 7⁴ = 2401 \(\therefore\) factors of 2401 and 1, 7, 49, 343 and 2401, i.e., 5 in number and so on.

Answer»

The number has to be resolved into prime factors. If the prime factorisation of the given number = 2^m × 3ⁿ × 5^p × 11^q , then the total number of factors = (m + 1) × (n + 1) × (p + 1) × (q + 1).

For example, if N = 2⁴ × 5⁶× 7⁴ × 13¹ , then

Number of factors of N = (4 + 1) × (6 + 1) × (4 + 1) × (1 + 1) = 5 × 7 × 5 × 2 = 350.

2⁴ = 16

\(\therefore\) factors of 16 are 1, 2, 4, 8, 16, i.e., 5 in number, i.e., (4 + 1) in numbers.

Similarly 7⁴ = 2401

\(\therefore\) factors of 2401 and 1, 7, 49, 343 and 2401, i.e., 5 in number and so on.

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