nb of divisors ?

[francoisph]

How many integral divisors does the number 120 have?

thks

[Rich@VeritasPrep]

Check out this post I wrote for a hint

https://blog.knewton.com/2010/05/06/gmat ... d-factors/

[jube]

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

[mj78ind]

Check out this post I wrote for a hint

https://blog.knewton.com/2010/05/06/gmat ... d-factors/

This is a cute little trick I picked up on the BTG forum thanks Raz!

[francoisph]

How to get quickly that 120 = 2^3 x 3 x 5 ??

120=(2^3)(3)(5)

[Rich@VeritasPrep]

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!

[amising6]

How many integral divisors does the number 120 have?

t 120=2*2*2*3*5
integral divisor 2^3*3*5
(3+1)(1+1)(1+1)=16

[francoisph]

thks raz
I really need to practice prime trees !

o:)