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Bunuel

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.


a = 2^4 * b^1 where b is a prime <> 2

Number of factors is (4+1)(1+1) = 10.

Answer E
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The question asks for DISTINCT factors, hence, they should be 1, 2, b and 16. Could someone explain why are we taking the total number of factors instead of the total number of distinct factors which is 4?
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The question asks for DISTINCT factors, hence, they should be 1, 2, b and 16. Could someone explain why are we taking the total number of factors instead of the total number of distinct factors which is 4?

There is no difference between those two: total number of factors and total number of distinct factors are the same.

Say, b = 3, then a = 16b = 2^4*3. The number of factors = (4 + 1)(1 + 1) = 10. The factors are 1 | 2 | 3 | 4 | 6 | 8 | 12 | 16 | 24 | 48 (10 divisors).
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Why cant b take a power of 0 then also it will be a prime number right ?
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C_K
Why cant b take a power of 0 then also it will be a prime number right ?
b is a prime number greater than 2; so b can be 3, 5, 7, 11, 13,.....or any other prime number. When you select any prime number, it will always have a power of 1. If b has power of 0, then b becomes 1, and 1 is not a prime number.
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