# K|n!+k from k=2,3,4 .n

1. Jul 8, 2014

### jonroberts74

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

prove by induction that K|n!+k for all integers $$n \ge 2$$ and k = 2,3,4,......,n

this one is different than other proofs I have had because Ive only dealt with n changing but now n and k change.

k goes up to n, does this mean k and n have the same values?

for instance,

P(2)

$$2|2! + 2 \Rightarrow 2|4$$

so p(k)

is $$k|n!+k; n=k$$

$$k|k!+k$$

??

Last edited: Jul 8, 2014
2. Jul 8, 2014

### Ray Vickson

You ask: "does this mean k and n have the same values?". No, that is not what it says and not what it means. Look at some small-n examples. If n = 3 it says that 2|3!+2 and 3|3!+3. So, k is not 3; it is any number in the set {2,3}. If n = 4 it says that all three of the statements 2|4!+2, 3|4!+3 and 4|4!+4 are true.

3. Jul 8, 2014

### jonroberts74

ah okay

whatever n is, then k takes on each value in the set {2,...,n}

4. Jul 8, 2014

### HallsofIvy

Staff Emeritus
The fact that if k< n the n k divides n! makes this pretty close to trivial.

5. Jul 8, 2014

### jonroberts74

$$2|n!+2$$
$$3|n!+3$$
$$4|n!+4$$
$$\downarrow$$
$$k|n!+k$$

where k takes on the values of 2,3,4...n and n is an integer greater than or equal to 2

and because for all k<n, k|n and when n=k, k|n

so my p(k) is

$$k|n!+k$$

and

p(k+1)

$$(k+1)|n!+(k+1)$$

Last edited: Jul 9, 2014
6. Jul 9, 2014

### jonroberts74

I have to make a correction. its not asking for induction. all the other problems on this problem set were induction but I confirmed with my prof this is not an induction proof question

7. Jul 9, 2014

### WWGD

Can you see a common factor in the numerator?

8. Jul 9, 2014

### jonroberts74

k is a factor of k, clearly and k is a also a factor of n! by the given information

$$2 \le k \le n$$ and by definition of a factorial k had to be used somewhere along the product

$$\prod_{k=2}^{n} k$$

case 1

k = 2, n > or equal 2

$$\frac{2*.....*n + 2}{2}$$

case 2

2 < k <n

$$\frac{2*...*k*....*n + k}{k}$$

case 3

n=k

$$\frac{2*......*k + k}{k}$$

Last edited: Jul 9, 2014
9. Jul 9, 2014

### HallsofIvy

Staff Emeritus
Those case are not necessary. WWDG's point is that if $k\le n$ then k is a factor of n!.

10. Jul 9, 2014

### jonroberts74

I understand that but I am expected to show a proof. Seems a tad bit too easy to say if

$$k \le n$$ then k is a factor of n!

11. Jul 9, 2014

### Ray Vickson

But that is absolutely correct, so IS a proof! Sometimes proofs are easy---after you "see" them.