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Finding Factors of Large Nums [#permalink]
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02 Mar 2010, 13:21
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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/triedandtrue strategy for finding factors of large numbers without having to create a chart, which seems too time consuming. Thanks in advance for the help!





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Re: Finding Factors of Large Nums [#permalink]
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02 Mar 2010, 18:16
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amenewyork wrote: 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/triedandtrue 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: 2^0*7^0 = 1*1 = 1 2^0*7^1 = 1*7 = 7 2^0*7^2 = 1*49= 49 2^1*7^0 = 2*1=2 2^1*7^1 = 2*7=14 2^1*7^2 = 2*49 = 98 2^2*7^0 = 4*1=4 2^2*7^1 = 4*7=28 2^2*7^2 = 4*49 = 196 thus, the factors of 196 are 1, 2, 4, 7, 14, 28, 49, 98, 196
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Re: Finding Factors of Large Nums [#permalink]
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04 Mar 2010, 08:56
Great, thank you for that method!



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Re: Finding Factors of Large Nums [#permalink]
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05 Mar 2010, 05:15
Yes the above method is the shortest method that i have come across...
Please share if you have come across anything shorter !



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Re: Finding Factors of Large Nums [#permalink]
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09 Mar 2010, 04:15
I agree. This is the shortest method.



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Re: Finding Factors of Large Nums [#permalink]
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09 Mar 2010, 14:16
Thanks for this method. And what about the number 483? We get three prime numbers 7, 23 and 3. What to do next?



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Re: Finding Factors of Large Nums [#permalink]
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10 Mar 2010, 03:16
amenewyork wrote: 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/triedandtrue 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 IntegerFirst 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). Hope it helps.
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Re: Finding Factors of Large Nums [#permalink]
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10 Mar 2010, 10:07
I meant how to find all 8 factors? As you did in your example: firasath wrote: amenewyork wrote: now work through the permutations of the exponents starting with 0:
2^0*7^0 = 1*1 = 1 2^0*7^1 = 1*7 = 7 2^0*7^2 = 1*49= 49
2^1*7^0 = 2*1=2 2^1*7^1 = 2*7=14 2^1*7^2 = 2*49 = 98
2^2*7^0 = 4*1=4 2^2*7^1 = 4*7=28 2^2*7^2 = 4*49 = 196
thus, the factors of 196 are 1, 2, 4, 7, 14, 28, 49, 98, 196
Does GMAT check this knowledge?



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Re: Finding Factors of Large Nums [#permalink]
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10 Mar 2010, 15:00
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fruit wrote: I meant how to find all 8 factors? As you did in your example: firasath wrote: amenewyork wrote: now work through the permutations of the exponents starting with 0:
2^0*7^0 = 1*1 = 1 2^0*7^1 = 1*7 = 7 2^0*7^2 = 1*49= 49
2^1*7^0 = 2*1=2 2^1*7^1 = 2*7=14 2^1*7^2 = 2*49 = 98
2^2*7^0 = 4*1=4 2^2*7^1 = 4*7=28 2^2*7^2 = 4*49 = 196
thus, the factors of 196 are 1, 2, 4, 7, 14, 28, 49, 98, 196
Does GMAT check this knowledge? 3^0*7^0*23^0 = 1 3^0*7^0*23^1 = 23 3^0*7^1*23^0 = 7 3^0*7^1*23^1 = 161 3^1*7^0*23^0 = 3 3^1*7^0*23^1 = 69 3^1*7^1*23^0 = 21 3^1*7^1*23^1 = 483 your factors are: 1,3,7,21,23,69,161,483
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Re: Finding Factors of Large Nums [#permalink]
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15 Mar 2010, 15:12
How do you quickly determine a,b, c and p, q, r in the typical n=a^p*b^q*c^r formula?
Seems like guesswork to me...




Re: Finding Factors of Large Nums
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