Problem 2 · An equation with six factors and 225
Six factors in arithmetic progression whose product must equal −225.
Integer answer, at most 4 digitsThe following equation has a unique integer solution. Which one is it?
$$(x-1)(x-3)(x-5)(x-7)(x-9)(x-11) + 225 = 0$$
Copa Cangur · SCM
Easy
Closed answer
Reasoned solution
Key idea: we need an integer $x$ with $(x-1)(x-3)\cdots(x-11) = -225$. Since $-225$ is odd, all six factors must be odd, i.e. $x$ must be even.
The six factors are six consecutive integers of the same parity (step $2$). Try the central value $x=6$:
$$5 \cdot 3 \cdot 1 \cdot (-1) \cdot (-3) \cdot (-5) = -225. \;\checkmark$$
To rule out the rest: if $x \le 0$ or $x \ge 12$, the product is positive (and very large), never $-225$. The remaining interior even values give:
$$x=2:\; -945, \qquad x=4:\; -315, \qquad x=8:\; +315, \qquad x=10:\; -945.$$
None equals $-225$. The integer solution is unique.
Answer: 6