Question

What is the greatest whole number that must be a divisor of the product of any three consecutive positive integers?

Answer 1

This deals with integers correct?

Answer 2

Yes and number theory to be precise

Answer 3

Let me see what i can do...

Answer 4

Hmm i am afraid my knowledge on this is to short... I am sorry @danyboy9169 I dont have much knowledge with gwh yet..

Answer 5

GWH=Greatest whole number.

Answer 6

Yes..I think the ans is 6 I got it right,, it is easy to do ..

Answer 7

Awesome nice to know you got it correct!

Answer 8

see if you take any three numbers and multiply them they are divisible by 6 which is the GWH.... I just did trial and error and got it right..thinking too deeply puts me on the wrong foot

Answer 9

Cool :)! I need to go help others now :)!

Answer 10

How to prove the lower bound is 6?

Answer 11

n(n-1)(n+1)

n(n^2-1)

n^3-n

seems to be all i got at the moment :)

Answer 12

The upper bound is 6 because 1*2*3=6.

Answer 13

1*2*3= 6
4*5*6= 120
6*7*8=336
Try any 3 consecutive #s they are always divisible by 6
We know that at least one of our three consecutive positive integers must be a multiple of $2$ since every other integer in a list of consecutive integers is divisible by $2$. Similarly, one of our three consecutive integers must also be divisible by $3$. Thus, the product of our three integers must be divisible by $2 \cdot 3 = 6$. By choosing the example where our three consecutive integers are $1$, $2$, and $3$ and their product is $6$, we see that $\boxed{6}$ is indeed the greatest whole number that must be a factor of the product of any three consecutive positive integers.

Answer 14

**\[n^{\phi(6)} \equiv 1 \pmod{6} \implies n^2\equiv 1 \pmod{6} \implies n^3\equiv n \pmod{6}\]

Answer 15

n(n+1)(n+2) -> Three numbers, one of them will be 0 mod 3.
n(n+1)(n+2) -> More than two numbers, at least one of them will be 0 mod 2.

Therefore, n(n+1)(n+2)=0 (mod 6)

Answer 16

The upper bound and lower bound are both 6. Therefore, the number is 6.

Answer 17

wow! great.. ganeshie8 and thanks Thomas5267, now it is more clear

Answer 18

Suddenly realised that the middle number between the twin primes must be divisible by 6.