Question

A stick of length $20$ units is to be divided into $n$ parts so that the product of the lengths of the parts is greater than unity.
The maximum possible value of $n$ is

• A. $20$
• B. $19$
• C. $18$
• D. $21$

Answer 1

Answer: (B)

$19$

Given, a stick is divided into $n$ parts which is of length $20$ units

Let the $n$ parts be $x_1,x_2,...,x_n$

$x_1+x_2+x_3+...+x_n=20$

Given, $x_1\ast x_2\ast ...\ast x_n>1$ -------(1)

We know that $A.M.\ge G.M.$

$A.M.$ of $x_1,x_2,...,x_n=\frac{x_1+x_2+...+x_n}{n}$

$G.M.$ of $x_1,x_2,...,x_n=\sqrt[n]{n{x}_1\ast x_2\ast ...\ast x_n}$

Therefore,

$\frac{x_1+x_2+...+x_n}{n}\ge \sqrt[n]{n{x}_1\ast x_2\ast ...\ast x_n}$

$\frac{20}{n}\ge \sqrt[n]{n{x}_1\ast x_2\ast ...\ast x_n}$ (since, $x_1+x_2+x_3+...+x_n=20$ )

taking $n^{th}$ power on both sides.

$(\frac{20}{n}{)}^n\ge x_1\ast x_2\ast ...\ast x_n$ -------(2)

From (1) and (2) we get

$(\frac{20}{n}{)}^n>1$

$⟹\frac{20}{n}>1$ (since, $\sqrt[n]{n1}=1$)

$n<20$

Therefore, the maximum possible value of $n=19$