(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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Prime factorization
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.
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.
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.
Hi nikhilgmat31,nikhilgmat31 wrote:there is no possibility of prime number in 12!13!
so 61 is answer.
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.
Cheers,
Brent
Here's a similar question to practice with: https://www.beatthegmat.com/p-12-11-t279341.html
Cheers,
Brent
Cheers,
Brent
I understand how to solve up to where we replace 14! (12! * 13 * 14) to the original equation.shovan85 wrote:Sure!! But please make sure you have a clear concept on factorials. See my previous post (for Quick look) and refer a book.rdjlar wrote:Hi, could somebody break down the factorization step por favor?
(13!12!) + (13!14!) = (13!12!)(1 + 14*13)
(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)
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!













