0
$\begingroup$

Suppose $n$ is a positive integer and $3$ arbitrary numbers are choosen from the set $\{ 1,2,3, \cdots , 3n+1 \}$ with their sum equal to $3n+1$. What is the largest possible product of those $3$ number ?

$\endgroup$
2
$\begingroup$

HINT Think about AM-GM. You get penalized more if you are away from the optimum in AM-GM.

Let the numbers be $a,b$ and $c$. From AM-GM, you have $$\dfrac{a+b+c}3 \geq \sqrt[3]{abc} \implies abc \leq \left(\dfrac{3n+1}3 \right)^3$$where the optimum is attained when$a=b=c=\dfrac{3n+1}3 = n + \dfrac13$. For the numbers in your set, the optimum is attained when $a=b=n$ and $c=n+1$. Hence, $$\max\{abc\} = n^2(n+1)$$

$\endgroup$
0
$\begingroup$

If $a \ne b \ne c$ then what is the largest possible product of those $3$ number ?

I think may be $(n-1)n(n+2)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.