number property ,

[tanviet]

if n and k are positive intergers, is n divisible by 6

1, n=k(k+1)(k-1)
2, k-1 is multiple of 3

pls, help with this.

[DanaJ]

My guess is A.

1. n = (k-1)k(k+1)
The product of two consecutive numbers will always be even, since one of them HAS to be even. This means that (k-1)k is divisible by 2, so n will also be divisible by 2
Now, the product of three three consecutive numbers will always be divisible by 3 in much the same way as above: out of three consecutive numbers, one MUST be divisible by 3. Pick any three consecutive numbers to prove that. This is why n is also divisible by 3.

Now, since we've established that n is divisible by both 2 and 3, then n will be divisible by 2*3 = 6.

2. Since there is no info about the link between k and n, we can't say if n is or is not divisible by 6.

[tanviet]

how to prove that product of 3 consecutive is divided by 6 evenly?

[DeepakR]

n=(k-1)k(k+1) where n and k are positive integers.

If k=1 then n=0*1*2=0 which is neither positive or negative integer. Hence A will fail. Now using A and B we can say that they are divisible by 6 example (3,4,5) or (6,7,8) or (9,10,11) etc.

So i would go ahead with C.)

- Deepak

[krisraam]

n=(k-1)k(k+1) where n and k are positive integers.

If k=1 then n=0*1*2=0 which is neither positive or negative integer. Hence A will fail. Now using A and B we can say that they are divisible by 6 example (3,4,5) or (6,7,8) or (9,10,11) etc.

So i would go ahead with C.)

- Deepak

When k =1 n = 0(Which is not a positive integer). So can't be 1.

Answer is A.

Thanks
Raama

[Musiq]

The 2 number properties you will need are below:

If you take 2 consecutive integers, one of them MUST be ODD and the other MUST be EVEN.

The product of X consecutive integers must be divisible by X. This can be seen by induction:

The product of 2 consecutive integers is divisible by 2
The product of 3 consecutive integers is divisible by 3
The product of 4 consecutive integers is divisible by 4...so on and so forth.

Using both of these properties and recognizing that Statement 1 is really the product of 3 consecutive integers is key to the problem.

Finally, if you can divide by 2 AND by 3, then you MUST be able to divide by 6.

Therefore A is the answer.

[vittalgmat]

how to prove that product of 3 consecutive is divided by 6 evenly?

u can try plugging 3 consecutive number. .
Alternately.
For every group of 3 consecutive numbers there will be atleast 1 multiple of 3 and 1 even number. Product of these two will definetely be divisible by 6.

HT helps

[aj5105]

(A)

[Ian Stewart]

The 2 number properties you will need are below:

If you take 2 consecutive integers, one of them MUST be ODD and the other MUST be EVEN.

The product of X consecutive integers must be divisible by X. This can be seen by induction:

The product of 2 consecutive integers is divisible by 2
The product of 3 consecutive integers is divisible by 3
The product of 4 consecutive integers is divisible by 4...so on and so forth.

Using both of these properties and recognizing that Statement 1 is really the product of 3 consecutive integers is key to the problem.

Finally, if you can divide by 2 AND by 3, then you MUST be able to divide by 6.

Therefore A is the answer.

We can actually extend the logic above to make it even more useful:

*** The product of n consecutive integers is always divisible by n!

So, for example, if you multiply 5 consecutive integers, that product will always be divisible by 5! = 120.

It's not too hard to see why: multiples of 5 are five apart, so if you take any five consecutive integers, exactly one of them will be a multiple of five (try it!). Because multiples of four are four apart, either one or two of the five consecutive integers will be a multiple of four. You'll also have either one or two multiples of three, and at least one even number which is not a multiple of four, giving us factors of 5, 4, 3 and 2. So the product of five consecutive numbers is divisible by 5*4*3*2 = 5!. There's nothing special about n=5 here; this is true for any positive integer n.

And, of course, if you know this, then you can see immediately that S1 is sufficient in the problem in the original post.