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Add 1 to each exponent and multiply them: (1+1)(2+1)(1+1)=12

I did not follow the approach , why 're we adding 1 ? . I get confused with this type of problems . Where Can i find a clear explanation for this approach theoretically? Appreciate the help !

Add 1 to each exponent and multiply them: (1+1)(2+1)(1+1)=12

I did not follow the approach , why 're we adding 1 ? . I get confused with this type of problems . Where Can i find a clear explanation for this approach theoretically? Appreciate the help !

thanks

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 original question:

How many distinct integers are factors of 90?

A. 6 B. 8 C. 9 D. 10 E. 12

\(90=2*3^2*5\), which means that the number of factors of 90 is: \((1+1)(2+1)(1+1)=12\).

Just want to clear a concept that the formula to compute the no of factors of a number includes identical factors?

2^ 1 ∗3^ 2 ∗5^ 1 =90

therefore, for perfect square, we need to eliminate the identical factors?

Thanks in advance.

The number of factors that you get is only the distinct factors and does not count the identical factors in case of perfect square twice.. eg 36=2^2*3^2.. no of factors= 3*3=9.. they are 1,2,3,4,6,9,12,18,36.. hope it clears the doubt
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Just want to clear a concept that the formula to compute the no of factors of a number includes identical factors?

2^ 1 ∗3^ 2 ∗5^ 1 =90

therefore, for perfect square, we need to eliminate the identical factors?

Thanks in advance.

Please refer two distinct concepts to calculate factors of any Number

1) By writing the Number as product of two Integers

2) By prime factorization and multiplying the powers of Distinct prime after adding one in each power of Primes

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