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28 Nov 2016, 07:03
Kudos
Bookmarks
Hello,
MGMAT introduces in its Number Properties Strategy Guide the following general rule for finding the number of total factors of a certain number:
If a number with more than two prime factors has prime factorization \(a^x * b^y * c^z\) (where a,b, and c are all primes), then the number has \((x+1)*(y+1)*(z+1)\) different factors.
I am confused about this rule, because MGMAT introduces this rule in a example with a number that has only two prime factors.
Does this rule also works with only two prime factors in any case?
Thanks for helping!
MGMAT introduces in its Number Properties Strategy Guide the following general rule for finding the number of total factors of a certain number:
If a number with more than two prime factors has prime factorization \(a^x * b^y * c^z\) (where a,b, and c are all primes), then the number has \((x+1)*(y+1)*(z+1)\) different factors.
I am confused about this rule, because MGMAT introduces this rule in a example with a number that has only two prime factors.
Does this rule also works with only two prime factors in any case?
Thanks for helping!
28 Nov 2016, 07:17
Kudos
Bookmarks
Dansaw
Hello,
MGMAT introduces in its Number Properties Strategy Guide the following general rule for finding the number of total factors of a certain number:
If a number with more than two prime factors has prime factorization \(a^x * b^y * c^z\) (where a,b, and c are all primes), then the number has \((x+1)*(y+1)*(z+1)\) different factors.
I am confused about this rule, because MGMAT introduces this rule in a example with a number that has only two prime factors.
Does this rule also works with only two prime factors in any case?
Thanks for helping!
MGMAT introduces in its Number Properties Strategy Guide the following general rule for finding the number of total factors of a certain number:
If a number with more than two prime factors has prime factorization \(a^x * b^y * c^z\) (where a,b, and c are all primes), then the number has \((x+1)*(y+1)*(z+1)\) different factors.
I am confused about this rule, because MGMAT introduces this rule in a example with a number that has only two prime factors.
Does this rule also works with only two prime factors in any case?
Thanks for helping!
It works with a number with any number of primes. For example:
The number of factors of 2^5 is (5+1)
The number of factors of 2^5*3^2 is (5+1)(2+1)
The number of factors of 2^5*3^2*11^1 is (5+1)(2+1)(1+1)
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.
Check for more here: math-number-theory-88376.html
28 Nov 2016, 07:29
Kudos
Bookmarks
Dansaw
Hello,
MGMAT introduces in its Number Properties Strategy Guide the following general rule for finding the number of total factors of a certain number:
If a number with more than two prime factors has prime factorization \(a^x * b^y * c^z\) (where a,b, and c are all primes), then the number has \((x+1)*(y+1)*(z+1)\) different factors.
I am confused about this rule, because MGMAT introduces this rule in a example with a number that has only two prime factors.
Does this rule also works with only two prime factors in any case?
Thanks for helping!
MGMAT introduces in its Number Properties Strategy Guide the following general rule for finding the number of total factors of a certain number:
If a number with more than two prime factors has prime factorization \(a^x * b^y * c^z\) (where a,b, and c are all primes), then the number has \((x+1)*(y+1)*(z+1)\) different factors.
I am confused about this rule, because MGMAT introduces this rule in a example with a number that has only two prime factors.
Does this rule also works with only two prime factors in any case?
Thanks for helping!
It's easy to find out why on earth the number of different factors is \((x+1)(y+1)(z+1)\)
Note that \(n=a^x b^y c^z\) and any factor of \(n\) is a combination of \(a,b,c\) with their powers.
For example, with \(45=3^2 \times 5^1\), 45 has factor \(1=3^0 \times 5^0\), or 45 has factor \(9=3^2\times 5^0\)
Likewise, \(n=a^x b^y c^z\) will has factors like \(f=a^{k_1}b^{k_2}c^{k_3}\) and \(0\leq k_1 \leq x;\: 0\leq k_2\leq y;\: 0\leq k_3\leq z\)
Now let's count possible selections of \(f\)
\(k_1\) could be any integer value from 0 to \(x\), so there are \(x+1\) selections
\(k_2\) could be any integer value from 0 to \(y\), so there are \(y+1\) selections
\(k_3\) could be any integer value from 0 to \(z\), so there are \(z+1\) selections
The total possible selections are \((x+1)(y+1)(z+1)\)
That is how the formula is formed.
