Question

If $a,b,c$ are three consecutive integers between $-10$ and $+10$ (both inclusive), how many {integer values are possible for} \[ \frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}\,? \]

• A. 0
• B. 1
• C. 2
• D. 3
• E. 4

Hint:

With consecutive integers, set $n-1,n,n+1$. Reduce the expression to a single variable and use divisibility: if $\frac{2n^2+1}{3n}$ is integer, then $n\mid(2n^2+1)\Rightarrow n\mid 1\Rightarrow n=\pm1$.

Answer 1

The Correct Option is C

Step 1: Parametrize consecutive integers.
Let \(b=n\), so \(a=n-1\) and \(c=n+1\). The condition \(-10 \leq a < b < c \leq 10\) implies: \[ -9 \leq n \leq 9. \]

Step 2: Simplify the expression in terms of \(n\).
Compute each part: \[ \begin{aligned} a^3+b^3+c^3 &= (n-1)^3 + n^3 + (n+1)^3 \\ &= 3n^3 + 6n, \\[6pt] abc &= (n-1)n(n+1) = n(n^2-1) = n^3 - n, \\[6pt] a^3+b^3+c^3 + 3abc &= (3n^3+6n) + 3(n^3-n) \\ &= 6n^3 + 3n = 3n(2n^2+1), \\[6pt] (a+b+c)^2 &= (3n)^2 = 9n^2. \end{aligned} \] Hence the expression becomes: \[ \frac{a^3+b^3+c^3+3abc}{(a+b+c)^2} = \frac{3n(2n^2+1)}{9n^2} = \frac{2n^2+1}{3n}, \qquad n \neq 0. \]

Step 3: Integer condition.
For \[ \frac{2n^2+1}{3n} \] to be an integer, \(n\) must divide \(2n^2+1\). But \[ 2n^2+1 \equiv 1 \pmod{n} \;\;\Rightarrow\;\; n \mid 1 \;\;\Rightarrow\;\; n=\pm 1. \] For \(n=\pm 1\), the value is: \[ \frac{2(1)^2+1}{3(\pm 1)} = \frac{3}{\pm 3} = \pm 1. \] Both are valid since \(-9 \leq n \leq 9, \; n \neq 0\).

The expression can take exactly two integer values: \(\boxed{-1}\) and \(\boxed{1}\).

Therefore, the count is \(\boxed{2}\).