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Prime factorization

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by DanaJ » Thu Nov 18, 2010 1:07 pm
Source: Veritas Prep

What is the greatest prime factor of (13!12!) + (13!14!)?

A. 11
B. 13
C. 47
D. 61
E. 71

Hehe, really liked this one! Let's see if you can nail it! :)
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by David@VeritasPrep » Thu Nov 18, 2010 1:08 pm
For anyone who does not already know how to do a problem like this, I think the following explanation will be very helpful!

Official Answer:

Correct answer: (D)

Solution: To find the greatest prime factor of this expression, we first expand the expression, and then simplify it by factoring. (You may skip the expansion step if you find it unnecessary, but it is helpful in understanding how the numbers combine.) Combine the factorials in the first half of the expression so that they read 13 x 12^2 x 11^2 x 10^2...down to 1^2. Combine the factorials in the second half of the expressions so that they read 14 x 13^2 x 12^2 x 11^2 x 10^2...down to 1^2. To factor, divide each term by the largest number that the two terms have in common. After expanding the factorials, you can see that they share one 13 and the other squared terms from 12 to 1. Factor out (13)(12!)^2, which is equivalent to 13!12!. We are left with (13!12!)(1 + [13 x 14]). Now simplify the expression inside the parentheses: 13 x 14 = 182. 1 + 182 = 183. Our expression now reads (13!12!)(183). We are now ready to find the greatest prime factor. The question is whether 183 has a prime factor greater than 13. (You should suspect that it does; otherwise, the problem is too easy.) Using divisibility rules, check to see whether 183 is divisible by 3. Because 1+ 8 + 3 = 12, 183 is divisible by 3. 183 = 3 x 61. 61 is prime. Thus, the original expression can be rewritten as (12!13!)(3)(61). The largest prime factor of that expression is 61.
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by prasanthk » Thu Nov 18, 2010 1:15 pm
13!12!(1+13*14)
13!12!*183
13!12!*3*61

so answer is D
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by Rahul@gurome » Thu Nov 18, 2010 1:18 pm
DanaJ wrote:What is the greatest prime factor of (13!12!) + (13!14!)?

A. 11
B. 13
C. 47
D. 61
E. 71

Hehe, really liked this one! Let's see if you can nail it! :)
(13!12!) + (13!14!) = (13!12!)*(1 + 13!14!) = (13!12!)*(183) = (13!12!)*(3)*(61)

Thus, greatest prime factor of (13!12!) + (13!14!) is 61.

The correct answer is D.


If anyone needs more explanation: 13! = 1*2*3*...*12*13 => Greatest prime factor of 13! is 13. Similarly greatest prime factor of 12! is 11. Thus, greatest prime factor of (12!13!) is 13. So greatest prime factor of (13!12!)*(3)*(61) is 61!

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    by GMATGuruNY » Thu Nov 18, 2010 1:19 pm
    DanaJ wrote:Source: Veritas Prep

    What is the greatest prime factor of (13!12!) + (13!14!)?

    A. 11
    B. 13
    C. 47
    D. 61
    E. 71

    Hehe, really liked this one! Let's see if you can nail it! :)
    Whenever we're given terms -- things being added or subtracted -- one approach is to factor out the greatest common factor: the greatest factor that can be evenly divided into each term.

    (13!12!) + (13!14!)
    = (13!12!)(1 + 14*13)
    = 13!12!(183)
    =13!12!(3*61).

    Thus, the greatest prime factor is 61.

    The correct answer is D.
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    by sumayahlaura » Wed Dec 08, 2010 8:23 pm
    could someone please explain what ! means when I see it in this context, it's the first time i've come across it. Perhaps i'm not up to this chapter in my study guide!
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    by shovan85 » Wed Dec 08, 2010 10:32 pm
    sumayahrose wrote:could someone please explain what ! means when I see it in this context, it's the first time i've come across it. Perhaps i'm not up to this chapter in my study guide!
    Hi

    ! means Factorial.

    In mathematics, the factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n.

    For example,

    Say what is factorial of 5?

    We can write 5! = 1 * 2 * 3 * 4 * 5 = 120

    One exception 0! = 1 (not 0)

    For more Info: https://en.wikipedia.org/wiki/Factorial
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    by rdjlar » Thu Dec 09, 2010 7:07 am
    --
    Last edited by rdjlar on Thu Jul 14, 2011 8:53 am, edited 1 time in total.
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    by shovan85 » Thu Dec 09, 2010 8:58 am
    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)
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    by edvhou812 » Thu Dec 30, 2010 12:12 am
    I've been a member of this community for a little over a month, and I can't help but wonder if questions such as this can be solved within two minutes with pencil and paper. Some of the explanations from experts seem to include formulas that have been exported from Excel. Can someone please give me some insight?
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    by anshumishra » Thu Dec 30, 2010 1:59 am
    edvhou812 wrote:I've been a member of this community for a little over a month, and I can't help but wonder if questions such as this can be solved within two minutes with pencil and paper. Some of the explanations from experts seem to include formulas that have been exported from Excel. Can someone please give me some insight?
    This question can definitely be solved in a minute (2 minutes is way too more time). Make sure you understand the logic, then I am sure you can solve it in a minute as well.
    Thanks
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    by Taran » Sun Jan 09, 2011 8:20 am
    I agree such questions appear difficult. But if you attempt them with a cool head, they will not appear that difficult. This is provided your basic concepts are clear. This particular one can be solved within 2 min. I do not see any formula used to solve this. Just take out highest common factors out and it becomes easy to solve.
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    by jsnipes » Mon Oct 17, 2011 12:13 pm
    thanks for the explanations.

    i somehow got stuck at 13!(12!+14!) when the obvious next step is to factor out 12! and get 13!12!(1+[14*13])...13!12!(183) and 3*61=183. ergo 61 is answer.
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    by Tarun Khanna » Fri Oct 28, 2011 12:50 pm
    can someone explain from 3x61x12!x13! how can we deduce that 61 is the greatest prime factor...what about 13!x12!....we need not multiply this and check?? 61 is the greatest prime factor of 183 but how have we deduced that it is the greatest prime factor of 3x61x12!x13! ??
    what am i missing here ??

    thanks
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    by GmatMathPro » Fri Oct 28, 2011 4:03 pm
    Tarun Khanna wrote:can someone explain from 3x61x12!x13! how can we deduce that 61 is the greatest prime factor...what about 13!x12!....we need not multiply this and check?? 61 is the greatest prime factor of 183 but how have we deduced that it is the greatest prime factor of 3x61x12!x13! ??
    what am i missing here ??
    13!x12! cannot contain any prime factors greater than 13. Suppose we think it might contain a larger prime factor, let's say 17. Well, clearly 13! and 12! do not have any factors of 17 because 13 and 12 are both less than 17. So if there is a factor of 17, it must be the case that two numbers lower than 17, that are contained in 13!12! are multiplied together to create a 17. But this is impossible because 17 is prime. By definition, the only factors of a prime number are itself and one. So, no matter how big 13!*12! is (and it is huge), there is no way it can contain any factors of 17 or any other prime number greater than 13.

    Also, think about 13!*12! written out the long way. Each number in that product is either prime or can be broken down into a product of primes: 13*12*11*10*9...=13*(2*2*3)*11*(5*2)*(3*3).... Clearly no prime factors will be present that are bigger than 13, and when we multiply them all together, there is no way to create a bigger prime number because that would contradict the very definition of a prime number.
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