always divisible by 7!?

Question

my book says ,

(n-3)*(n-2)*(n-1)*n*(n+1)*(n+2)*(n+3)

is always divisible by 7!

is it correct ?

please comment on it .

Answer 1

(n-3) * (n-2) * (n-1) * n * (n+1) * (n+2) * (n+3) (assuming you meant to go to n+3) can be expressed as

(n+3)!/(n-4)!

Now let m = n+3 so then n - 4 = (m - 3) - 4 = m - 7. In terms of m we can write this as

m!/(m-7)!.

If this is to be divisble by 7! then [(m!/(m-7!)]/7! must be an integer. But

[(m!/(m-7!)]/7! = m!/((m-7)!*7!) = mC7, which is an integer and so (n+3)!/(n-4)! is divisible by 7!.

Math (and Stats) Rule!

Answer 2

Well asuming that the complete chain is:

(n-3) * (n-2) * (n-1) * n * (n+1) * (n+2) * (n+3)

to prove that it is always divisible by 7, you have to prove that at least one of the factors is divisible by 7.

Let's look at the problem a little closer. Multiples of seven go:

7, 14, 21, 28, 35, 42, etc etc

If you were to put this mutiples in the line of the real numbers, you will notice that between each pair, there are six integers.

between 7 and 14, (8, 9, 10, 11, 12, 13) and so on.
If n would be on the first position (in our example, 8), n-1, will be a factor of 7.
If n would be on the second position (in our example, 9), n-2, will be a factor of 7.
If n would be on the third position (in our example, 10), n-3, will be a factor of 7.
If n would be on the fourth position (in our example, 11), n+3, will be a factor of 7.
If n would be on the fifth position (in our example, 12), n+2, will be a factor of 7.
If n would be on the fourth position (in our example, 13), n+1, will be a factor of 7.
Of course n in any other position (in our example, 7 and 14) will be a multiple of 7 by itself.

If n=0 (or any factor is zero) it will still be divisible by 7.

Answer 3

That depends on how far up those ... go. If they went on forever, you'd have an infinite expression there.

I think what you want is:

(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3)

Based on this, I can explain why.

Among the seven factors, one must be divisible by 7 because one in every seven consecutive numbers is a multiple of 7.

By the same argument, one must be divisible by 6, and one by 5, and so on.

What's nice is that, for example, if we have the one divisible by 6 and then we want one divisible by 3, we don't want to choose that one again. However, we can go to one three away (either left or right) and pick that one. Same works when we have to pick the 2 - there's enough in there that one of them must be divisible by 2, but not by 4 or 6.

I'll work an example, n = 10

7*8*9*10*11*12*13

the 7 is divisible by 7
the 12 is divisible by 6
the 10 is divisible by 5
the 8 is divisible by 4
the 12 is divisible by 3 but we already used the 12
we can use the 9 instead though (see what I meant?)
and alot of them are divisible by 2. we've already used the 8 and 12 for the 4 and 6, but we can use the 10

You factor all those out and get:

7*6*5*4*3*2 (1)(3)(1)(11)(2)(13)

So we know that among them, one is divisible by 7, one is divisible by 6, one is divisible by 5, one is divisible by 4, one is divisible by 3 (not repeating the one divisible by 6), and one is divisible by 2 (not repeating the one divisible 6 or the one divisible by 4).

And so you're done, because that means that you can find all those factors separately, and factor them out, giving you a larger factor of 7*6*5*4*3*2 = 7!

Answer 4

I don't know how many terms you have in that product, because some of it was truncated.

But if you have the product of 7 consecutive nos., then YES it is always divisible by 7. It's because somewhere within those 7 numbers will be a multiple of 7. Thus the product will also be a multiple of 7.

**EDIT**
OK you meant 7 factorial. Check out Math Chick's answer below. It's spot on.

Answer 5

Of course it is correct. From 7 numbers in a sequence one must be divisible by 7.

Answer 6

A three-digit number with digits "abc" could be written as 100*a + 10*b + c. The six digit number then would be written as 100000*a + 10000*b + 1000*c + 100*a + 10*b + c. After factoring we obtain (100000 + 100)*a + (10000 + 10)*b + (1000 + 1)*c = 100100*a + 10010*b + 1001*c. Now factor again to obtain 1001*(100*a + 10*b + c). Of these two factors 1001 = 7* 143, so the original six-digit number will be divisible by 7 because one of its factors is divisible by 7.

Answer 7

Yes, the product of seven consecutive numbers
is always divisible by 7, because one of the numbers
(exactly one) must be a multiple of 7.

Answer 8

There is a simpler way to know whether any number is exactly divisible by 2,3,4.....till 12 with no remainders. Check up on sources which give you the rules on divisibility.Just takes about 2 to 3 mins to determine. May come back once i get hold of them.

Answer 9

49

Answer 10

i don't think so
how can an equation like this fit for a number divisible by 7.