Consecutive Shouting

Number Theory Level 1

$\large n!=\frac{2013!+2014!}{2015}, \ \ \ \ \ n = \ ?$

The answer is 2013.

6 solutions

Danish Ahmed
Apr 19, 2015

$n! = \dfrac{2013!(1 + 2014)}{2015} = \dfrac{2013!\cdot2015}{2015}=2013!$

So $n=2013$

Moderator note:

Good. Can you generalize this?

Here's a generalized form:

$\forall~n,a,m\in\Bbb{Z^+}~,~~n!=\frac{a!+(a+m)!}{1+\displaystyle\prod_{k=1}^m(a+k)}\implies n=a$

For the problem here, take $a=2013~,~m=1$ .

Prasun Biswas - 6 years, 1 month ago

Can you plz explain how to generalize ? I really love to know this stuff.

Syed Baqir - 5 years, 10 months ago

Wow, this is fun!

Jon Adams - 6 years, 1 month ago

Dhiraj Sharma
Apr 21, 2015

perfect solution.. Gooooooooood

Asia MALIK - 5 years, 11 months ago

Jinu Sudheendran
Apr 21, 2015

Eric Beck
Mar 16, 2016

We know the permutation 2013! = 2013 × 2012 × 2011 × 2010 × 2009 × 2008 ... × 1 and we know 2014! = 2014 × 2013 × 2012 × 2011 × 2010 × 2009 × 2008 ... × 1 so because of our knowledge with algebra we can isolate a 2014 and multiply that by 2013! to get 2014 × 2013! which in its natural permutation is 2014!. Then, factor out as shown:

n! = 2013! ( 1 + 2014 ) --> Notice that 1 + 2014 = 2015 and 2015 is in the denominator of the original equation so the ( 1 + 2014 ) on top and the 2015 on bottom cancel out. This then leaves us with:

n! = 2013!

n = 2013

Edward Nirenberg
May 9, 2015

2015n!= 2013!+2014!

2015n!=2013!(2014+1)

2015n!=2013!(2015)

n!=2013!

n=2013

Vishakan Subramanian
May 7, 2015

This actually works out by that taking out the common factor method, as well as other numbers.... like.... x!+ (x+1)!/(x+2) = x! I tried this out with 3,4,5 and 4,5,6 ....it was right ....

Consecutive Shouting

The answer is 2013.

6 solutions

Moderator note:

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