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Hi everyone, I'm beginning my GMAT Quant journey with the MGMAT Number Properties guide and the Kaplan Math Workbook (6th ed.). I understand the concept of factoring, finding GCFs, etc, but I'm wondering if there is a quick/tried-and-true strategy for finding factors of large numbers without having to create a chart, which seems too time consuming. Thanks in advance for the help!

Hi everyone, I'm beginning my GMAT Quant journey with the MGMAT Number Properties guide and the Kaplan Math Workbook (6th ed.). I understand the concept of factoring, finding GCFs, etc, but I'm wondering if there is a quick/tried-and-true strategy for finding factors of large numbers without having to create a chart, which seems too time consuming. Thanks in advance for the help!

Not sure what your "chart" method is but an easy method to remember and use is to find the prime factorization and then expand the permutations of the exponents:

example:

196

196 = 2^2*7^2 (prime factorization)

To determine how many factors there are, just look at each power, add one, and multiply together.

(2 + 1)(2 + 1) (3)(3) 9 factors

now work through the permutations of the exponents starting with 0:

Hi everyone, I'm beginning my GMAT Quant journey with the MGMAT Number Properties guide and the Kaplan Math Workbook (6th ed.). I understand the concept of factoring, finding GCFs, etc, but I'm wondering if there is a quick/tried-and-true strategy for finding factors of large numbers without having to create a chart, which seems too time consuming. Thanks in advance for the help!

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.

fruit wrote:

Thanks for this method. And what about the number 483? We get three prime numbers 7, 23 and 3. What to do next?

\(483=3^1*7^1*23^1\) --> number of factors of 483 is: \((1+1)(1+1)(1+1)=8\).

For more on number properties see the Number Theory chapter of Math Book (link in my signature).

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