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Prime factorization

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by Java_85 » Wed Sep 11, 2013 7:06 am
Nice Question. It took me more than 3 minutes!
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by [email protected] » Tue Nov 26, 2013 3:38 am
(13!12!) + (13!14!)

We take common hence we get (13!12!)(1+13*14)= 13!*12!*183

To further simplify we get 13*12*61*3 we can clearly see 61 is the greatest prime factor!
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by sahilbilga » Sat Jan 25, 2014 1:08 am
I think the answer is 61. option D. It (12!.13! + 12!.14!) = 12!.13!(1+182). So we have to find the largest prime factor of 183 from the given numbers and it is 61.
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by jaspreetsra » Mon Nov 03, 2014 2:13 pm
D
Explanation:
(13!12!) + (13!14!)
=(13!12!)(1+14*13)
=(13!12!)(1+182)
=(13!12!)(183)
=(13!12!)(61*3)
So, answer is 61.
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by nikhilgmat31 » Wed Jul 01, 2015 12:27 am
there is no possibility of prime number in 12!13!

so 61 is answer.
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by Brent@GMATPrepNow » Wed Jul 01, 2015 5:51 am
nikhilgmat31 wrote:there is no possibility of prime number in 12!13!

so 61 is answer.
Hi nikhilgmat31,

I'm trying to follow your logic above.
With the exception of 1, all integers will have at least one prime factor.
For instance, 12!13! has 2, 3, 5, 7, 11 and 13 as prime factors.

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by Brent@GMATPrepNow » Wed Jul 01, 2015 5:53 am
Here's a similar question to practice with: https://www.beatthegmat.com/p-12-11-t279341.html


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by nikhilgmat31 » Wed Jul 01, 2015 9:27 pm
yes Brent there are other prime factors, What I mean is , all are less then 61.
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by jo0sunee » Mon Sep 14, 2015 8:29 am
shovan85 wrote:
rdjlar wrote:Hi, could somebody break down the factorization step por favor?

(13!12!) + (13!14!) = (13!12!)(1 + 14*13)
Sure!! But please make sure you have a clear concept on factorials. See my previous post (for Quick look) and refer a book.

(13!12!) + (13!14!)

= (13!12!) + (13!14!) Just concentrate on what is 14!

14! = 1*2*3*.....*14 (Multiplication of all intgers starting from 1 to 14)

=> 14! = (1*2*3*...*12)*13*14 (Now I have selected from 1 to 12 in the multiplication list)

=> 14! = 12! * 13*14 (Multiplication of 1 to 12 is 12!)

Now put this value of 14! in our actual question.

(13!12!) + (13!12! * 13*14)

Thus we can take common 13! 12! so, (13!12!) (1 + 14*13)
I understand how to solve up to where we replace 14! (12! * 13 * 14) to the original equation.
What I don't understand is how 14! changes the latter half of equation to (1 + 14*13), and what occurs after as well.

Please help!
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