Solve question 39 ?

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2 Answers
Jan 10, 2018

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.

Jan 10, 2018

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