Question

Solve question 39 ?

Answer 1

B

Explanation:

First, we should make use of the fact that the numbers must be consecutive, by calling the numbers we choose to be `n-1,n,n+1`, where if we abide by the constraints `n` must be between `-9` and `9` inclusive.

Second, notice that if we get a certain value for a specific `a,b,c`, we can swap around those specific values, but still get the same result. (I believe this is called being permutable but forget the proper term)

So we can simply let `a=n-1`,`b=n`,`c=n+1`, now we plug this in:

`(a^3+b^3+c^3+3abc)/(a+b+c)^2`
`=((n-1)^3+n^3+(n+1)^3+3(n-1)(n)(n+1))/(n-1+n+n+1)^2`
`=(n^3-3n^2+3n-1+n^3+n^3+3n^2+3n+1+3n(n^2-1))/(3n)^2`
`=(n^3+3n+n^3+n^3+3n+3n^3-3)/(9n^2)`
`=(6n^3+6n-3)/(9n^2)`
`=(2n^3+2n-1)/(3n^2)`

Now our problem becomes to see for what values of `-9<=n<=9` the expression gives an integer values, how many different values we get.

I'm going to continue the solution in a separate answer just to make it easier to read.

Answer 2

Part 2 of my sol'n. This will be using modular arithmetic, but if you are unfamiliar with it then there is always the option of subbing in all necessary values of `n`

Explanation:

Because the expression must be an integer value, the bottom must divide the top exactly. Thus, the numerator should have a factor of 3. And for this we should use modular arithmetic.

Examine for which n satisfies: `2n^3+2n-1-=0 mod3`
`2n^3+2n-=1 mod3`
`2n^3+2n-=-2 mod3`
`n^3+n-=-1 mod3`
Now casework:
1. We try `n=3k`
`LHS=(3k)^3+3k`
`=3(9k^3+k)-=0 mod3`, which doesn't work
2. We try `n=3k+1`
`LHS=(3k+1)^3+(3k+1)`
`=(3k+1)^3+(3k+1)`
`=27k^3+27k^2+27k+1+3k+1`
`-=2-=-1 mod3`, which works
3. We try `n=3k-1`:
`LHS=(3k-1)^3+(3k-1)`
`=27k^3-27k^2+27k-1+3k-1`
`-=-2-=1`, which doesn't work

So we deduce that `n` must be of the form `3k+1`, or one more than a multiple of 3. Considering our range for n, being `-9<=n<=9`, we have the possible values of:
`n=-8,-5,-2,1,4,7`.

At this point you might able to use the fact that `n=3k+1`, but with only 6 values to check I decided to instead calculate each one instead, and the only value for `n` that works is `n=1`, producing the result of `1`.

So finally, the only set of consecutive numbers that produces an integer result is `0,1,2`, giving `1` hence the answer is `B`