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francoisph
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https://blog.knewton.com/2010/05/06/gmat ... d-factors/
nb of divisors ?
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Source: Beat The GMAT — Problem Solving |
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Rich@VeritasPrep
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Yep! It's exactly like Rich said!
If n=(p1^a)(p2^b)(p3^c).... where p1, p2, p3 are the prime factors for n & a, b, c... are the powers of that particular prime no. in that number, then the no. of factors (or divisors) of n=(a+1)(b+1)(c+1)....
for e.g. 28=(2^2)7. Therefore no. of factors for 18=(2+1)(1+1)=6
120=(2^3)(3)(5)=(3+1)(1+1)(1+1)=16
If n=(p1^a)(p2^b)(p3^c).... where p1, p2, p3 are the prime factors for n & a, b, c... are the powers of that particular prime no. in that number, then the no. of factors (or divisors) of n=(a+1)(b+1)(c+1)....
for e.g. 28=(2^2)7. Therefore no. of factors for 18=(2+1)(1+1)=6
120=(2^3)(3)(5)=(3+1)(1+1)(1+1)=16
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mj78ind
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This is a cute little trick I picked up on the BTG forum thanks Raz!raz1024 wrote:Check out this post I wrote for a hint
https://blog.knewton.com/2010/05/06/gmat ... d-factors/
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francoisph
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Rich@VeritasPrep
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Practice those factor trees!
You can break down 120 pretty quickly just by starting with an obvious divisor, namely 2.
120
2 * 60
60 is another multiple of 2, so keep going with that:
2 * 2 * 30
Another multiple of 2, so keep going!
2 * 2 * 2 * 15
15 is pretty easy to break down into 3 and 5:
2 * 2 * 2 * 3 * 5
Done!
You can break down 120 pretty quickly just by starting with an obvious divisor, namely 2.
120
2 * 60
60 is another multiple of 2, so keep going with that:
2 * 2 * 30
Another multiple of 2, so keep going!
2 * 2 * 2 * 15
15 is pretty easy to break down into 3 and 5:
2 * 2 * 2 * 3 * 5
Done!
Rich Zwelling
GMAT Instructor, Veritas Prep
GMAT Instructor, Veritas Prep
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francoisph
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