# Number Theory non zero natural numbers

1. Feb 23, 2013

### mtayab1994

1. The problem statement, all variables and given/known data

For all non zero natural numbers n prove that:

1- $$24\mid n(n+1)(n+2)(n+3)(n+4)$$

and that :

2- $$120\mid n(n+1)(n+2)(n+3)(n+4)(n+5)$$

3. The attempt at a solution

1- For n=1 we get that 24 divides 120 so we assume that 24 divides n(n+1)(n+2)(n+3)(n+4)

and we show that 24 divides (n+1)(n+2)(n+3)(n+4)(n+5). From the first step we have that

24 divides (n+1)(n+2)(n+3)(n+4), I can't quite finish it off. Any help would be appreciated.

2. Feb 23, 2013

### micromass

Staff Emeritus
Induction is not needed.

Can you prove that 3 always divides 3 subsequent numbers??

Now we have to prove that 24 always divides 5 subsequent numbers. Think about the prime factorization of 24.

3. Feb 23, 2013

### mtayab1994

If 3 divides 3 subsequent numbers than 3 either divides n or n+1 or n+2

4. Feb 23, 2013

### micromass

Staff Emeritus
Yes, true. But it is clear why 3 always divides 3 subsequent numbers?

5. Feb 23, 2013

### mtayab1994

I'm not pretty sure but i believe it is that when 3 subsequent numbers are multiplied together they are always in the from 3k(3k+1)(3k+1) and making that 3k is always divisible by 3.

6. Feb 23, 2013

### micromass

Staff Emeritus
Good. Now perhaps simplify the question in your OP by splitting 24 up in factors.

7. Feb 23, 2013

### mtayab1994

Ok when splitting up 24 into factors i got that : $$24=2^{3}*3$$

And 3 divides n(n+1)(n+2)(n+3)(n+4) because 3 divides n(n+1)(n+2) and 2^3 divides (n+1)(n+2)(n+3)(n+4) because 2^3 divides n(n+1)(n+2)(n+3) and 3 and 8 are prime to each other so 3*2^3 divides n(n+1)(n+2)(n+3)(n+4) right?

Last edited: Feb 23, 2013
8. Feb 23, 2013

### micromass

Staff Emeritus
Yes. Go on... Do the factors individually need to divide the number?

9. Feb 23, 2013

### mtayab1994

Sorry i edited my previous post while you were replying.

10. Mar 6, 2013

### Joffan

The problem appears to be on the unambitious side. Actually, for $n \in \mathbb{N}_1$,

a) $24\text{ | } n(n+1)(n+2)(n+3)$

and

b) $120 \text{ | } n(n+1)(n+2)(n+3)(n+4)$

Think modularly...