happyface101 Wrote:Hi Experts,
There are a few "rules" below that are deemed to be true, and after testing a few examples, I know they are true. However, can someone please explain / prove why they are true (other than testing a few examples - I have done this)?
It's hard for me to just memorize rules without understanding why they are true. I think a lot of other people are the same too, so this could be helpful for everyone. Thanks in advance!
1. "The product of n consecutive integers is always divisible by n! So, 4x5x6 (4*5*6=120) is divisible by 3!"
2. "If you have an odd number of terms in consecutive set, the sum of those numbers is divisible by the number of terms. This does not hold true for consecutive sets with an even amount of terms."
3. "If the arithmetic mean of 3 consecutive integers is odd, the product of them is divisible by 8"
Hi,
I am not really an expert but hope you do not mind if I take a shot at this and hopefully an expert can enlighten us both.
> I think hold all these hold true for only positive integers.
1) n consecutive integers are by definition equally spaced apart with a difference of 1. So in simple terms n integers will be of the order n, n+1,n+2....
Factorial of n, by definition is the product of all non zero numbers right up to n i.e n(n-1)(n-2)....1.
These consecutive numbers are equally spaced apart - the difference being one. Now for divisibility, it is essential that the we have common factors that cancel each other out - that is to say the numerator must have "at least" the same factors as the denominator. The fact that we are dividing by n!, supplies that least number of factors. If we consider the example that you have stated , we realize that any number which has 3 consecutive integers will "always" be divisible by 3, or generally every n consecutive integers will be divisible by n [ reason - think of the multiples of 3, they come in intervals of 3] - so one factor supplied . Now for the remaining - we know that 3 consecutive numbers will "always" have one even number and every factorial will have at most half of its numbers even, so even factors are now supplied . Hence we have a winner!!
2) True - Let there be 3 numbers n,n+1,n+2. Add them - 3(n+1) i.e divisible by 3 (the number of terms) - I get n+1 . If I have 4 (even) integers , add them- 4n+6, now try an divide by 4 - I get a non integer, We can see clearly that breaking down 4n + 6 / 4 creates a n + 2/3 - an integer plus a fraction, so not a clean division here!
Extend this test to any subset of numbers and you will realize that the rule that you have stated is true.
3) True - Let there be 3 positive integers n,n+1,n+2. Add them - 3(n+1) and divide by 3 (the number of terms) to get the mean - n+1, for the value n+1 to be odd, n
must be even so let us restate the question as "If n is even, the product of n(n+1) (n+2) is divisible by 8". now take the minimum possible even value of n=2 an solve , 2 * 3 * 4 / 8 =3.
Extend this test to any subset of numbers and you will realize that the rule that you have stated is true - basically what happens here is this we have 2 evens(n and n+2) and one odd (n+1) , product of the 3 is even , When we took 2 in the equation as n , the third number was 2+2 = 4 so 2*3*4 has enough factors to offset the 2*2*2 of 8. Since the minimum even number solves this , consider that a greater even number (n and n+2) will supply a greater number of 2's and therefore easily cancel 8 out. Hence True!
> From my understanding of what Ron says , a grasp of basic number properties will help much more than a memory full of formulae. It is okay to memorize formulae in the end but only after you understand how odds, evens, positive etc. negatives work.
Thanks