# Proving this

1. Sep 13, 2004

### decibel

is there a way to prove that n consecutive integers is always divisible by n!?

2. Sep 13, 2004

### Gokul43201

Staff Emeritus
I think you have worded this question incorrectly. Going strictly by your wording, the answer would be : No...in fact, there are no two consecutive numbers that are divisible by any given number ( >1).

I think I know what you might be intending to ask, but I'll let you pose the question correctly, first.

3. Sep 13, 2004

### decibel

sorry, but i dont know what your saying, where did two come from?

4. Sep 13, 2004

### Gokul43201

Staff Emeritus
If no two consecutive integers are divisible by any integer ( >1), how can you find n consecutive integers ?

5. Sep 13, 2004

### Alkatran

starting equations:
n mod y = 0
(n+1) mod y = 0

work:
((n mod y) + (1 mod y)) mod y = 0
(n mod y) + 1 = 0
(n mod y) + 1 = n mod y
1 = (n mod y) - (n mod y)
1 = 0

conclusion:
FALSE

No two consecutive integers are divisible by the same factor.

6. Sep 13, 2004

### phoenixthoth

Easier: let n=3. The statement is that 3 consequtive integers is divisible by 6. Well, 1,2,3 provides a counterexample.

Wait. I'm not sure what you mean now. Do you mean the SUM of n consequtive integers is divisible by n!?

If that's what you meant, then 2, 3, 4 has sum 9 and is not divisible by 6(=3!).

7. Sep 13, 2004

### Gokul43201

Staff Emeritus
The more common question is : Prove that you can always find n consecutive composite numbers.

Of course, these are be (n+1)!+2, (n+1)!+3, (n+1)!+4, ...(n+1)!+(n+1), and the proof is quite self-evident.

8. Sep 13, 2004

### Gokul43201

Staff Emeritus
Just to make this complete (not to nitpick)

starting equations:
n mod y = 0
(n+1) mod y = 0

work:
((n mod y) + (1 mod y)) mod y = 0
(n mod y) + 1 = 0 {y <> +/- 1}
(n mod y) + 1 = n mod y
1 = (n mod y) - (n mod y)
1 = 0

conclusion:
FALSE

No two consecutive integers are divisible by the same factor (>1).

9. Sep 13, 2004

### Alkatran

I was thinking about that when I wrote it. I got lost on a tangent of what the remainder of mod 0 was... undefined, bah.

Gimme a break, it's the second time I try to prove something using modulo.

I have a question though, is:
(x op y) mod z == ((x mod z) op (y mod z)) mod z
true? (op being +, -, *, / , ^, etc) is it true for functions? (sin, f())

10. Sep 13, 2004

### Gokul43201

Staff Emeritus
Sorry 'bout that.

It is true, however for +, - and *.

I'm not sure how your question applies to functions like sin(x), when you are asking about binary operators.

11. Sep 13, 2004

### Alkatran

So, could I say:

If x op y = y op x then (x op y) mod z = ((x mod z) op (y mod z)) mod z

Alright, the sin() thing won't work, I was off on a tangent. What about binary operations on the numbers? (XOR, OR, AND)

12. Sep 13, 2004

### Gokul43201

Staff Emeritus
To clarify, when I said 'binary operator', I meant an operator defined on 2 numbers (such as addition, subtraction, exponentiation, etc), not operators defined on the binary representation of numbers.

I'm pretty sure that rule doesn't work for AND, OR, XOR, etc...but I haven't verified this.

"If x op y = y op x then (x op y) mod z = ((x mod z) op (y mod z)) mod z"

In general, this is not necessarily true.

Last edited: Sep 13, 2004
13. Sep 13, 2004

### Alkatran

I understand what you meant by binary operators. It just made me think of the logical operations. (computer programmer here...)

Do you have an example or shall I find one myself?
Got it:
5 and 7 = 5
5 mod 3 = 2
(5 and 7) mod 3 = 2
((5 mod 3) and (7 mod 3)) mod 3 = (2 and 1) mod 3 = 0 mod 3 = 0
0 <> 2

Last edited: Sep 13, 2004
14. Sep 13, 2004

### fourier jr

i know there's a theorem similar to what decibel asked about in lozansky's "winning solutions". i can't remember exactly how it went though, but it definitely had something about n consecutive integers & n!