Sometimes?

Number Theory Level 1

1 ! + 2 ! 1 ! − 2 ! ∈ Z \dfrac{1!+2!}{1!-2!} \in \mathbb{Z}

2 ! + 3 ! 2 ! − 3 ! ∈ Z \dfrac{2!+3!}{2!-3!} \in \mathbb{Z}

3 ! + 4 ! 3 ! − 4 ! ∉ Z \dfrac{3!+4!}{3!-4!} \not \in \mathbb{Z}

4 ! + 5 ! 4 ! − 5 ! ∉ Z \dfrac{4!+5!}{4!-5!} \not \in \mathbb{Z}

True or false:

m ! + n ! m ! − n ! ∉ Z \dfrac{m!+n!}{m!-n!} \not \in \mathbb{Z} for every consecutive integers ( m , n ) (m,n) . With m > 2 , n > 2 m>2, n>2 and m < n m<n .

False True

Munem Shahriar by Munem Shahriar , 18, Bangladesh

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Plinio Sd
Feb 2, 2018

Indeed, m ! + n ! m ! − n ! ∉ Z \frac{m!+n!}{m!-n!}\notin\mathbb{Z} for any pair of integers ( m , n ) (m,n) , 2 < m < n 2< m<n . Let k = n ! / m ! k = n!/m! , then m ! + n ! m ! − n ! = m ! m ! 1 + n ! / m ! 1 − n ! / m ! = − k + 1 k − 1 = − 1 − 2 k − 1 . \frac{m!+n!}{m!-n!} = \frac{m!}{m!}\frac{1+n!/m!}{1-n!/m!} = -\frac{k+1}{k-1} = -1 - \frac{2}{k-1}. The above expression is an integer only if k ∈ { − 1 , 0 , 2 , 3 } k\in\{-1,0,2,3\} , otherwise we would forcefully have a fraction. However, there are no values of n , m ∈ Z n,m\in\mathbb{Z} , 3 ≤ m < n 3\le m<n that satisfies this restriction, since n ! n! is at least 4 times greater than m ! m! .

0 Helpful 0 Interesting 0 Brilliant 1 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...