Let , N be roots of the equation x^2-70x+=0, where /2,/3 N.

Question

Let $\alpha,\beta\in\mathbb N$ be roots of the equation $x^2-70x+\lambda=0$, where $\lambda/2,\lambda/3
otin\mathbb N$. If $\lambda$ assumes the minimum possible value, then $|\alpha-\beta|$ is equal to:

Answer: 60

Accepted Answer

Since $\alpha+\beta=70$ and $\lambda=\alpha\beta$ must be odd (as $\lambda/2\notin\mathbb N$) and not a multiple of 3, both $\alpha,\beta$ must be odd, and neither can be a multiple of 3 (else $\lambda$ would be).

For a fixed sum, the product $\alpha\beta$ is smallest when $\alpha,\beta$ are as far apart as possible. Checking odd values of $\alpha$ not divisible by 3, from smallest upward:
- $\alpha=1\Rightarrow\beta=69$ (divisible by 3) --- rejected.
- $\alpha=5\Rightarrow\beta=65$ (neither divisible by 3) --- valid, $\lambda=5\times65=325$.

This is the minimum valid $\lambda$. So $\{\alpha,\beta\}=\{5,65\}$, and
$$|\alpha-\beta|=65-5=60.$$

Solutions · 1