Question

A stick of length $20$ units is to be divided into $n$ parts. So that the product of the length of the parts is greater than unity. The maximum possible of $n$ is
$
  a)\,20 \\
  b)\,19 \\
  c)\,18 \\
  d)\,21 \\
 $

Answer 1

Hint: Let a $20$ unit length be divided into $n$ parts thus length of each part may be ${x_1},{x_2},{x_3},........,{x_n}$
So ${x_1} + {x_2} + {x_3} + ........ + {x_n} = 20$ and given
${x_1}{x_2}{x_3}........{x_n} > 1$
And it is known that AM$ \geqslant $GM
Arithmetic mean is greater than Geometric mean.

Complete step-by-step answer:
Here it is given a stick of length $20$ units and it is divided into $n$ parts.
Now it is not given $n$ equal parts.
Divided into $n$ parts with unequal length
Let the length of each part may be
${x_1},{x_2},{x_3},........,{x_n}$
So ${x_1} + {x_2} + {x_3} + ........ + {x_n} = 20$
And it is given in the question the product of length of parts greater than unity.
So ${x_1}{x_2}{x_3}........{x_n} > 1$
And it is known that AM$ \geqslant $GM
Arithmetic mean is always greater than geometric mean.
Let us find the arithmetic means of ${x_1},{x_2},{x_3},........,{x_n}$
AM$ = \dfrac{{{x_1} + {x_2} + {x_3} + ........ + {x_n}}}{n}$
Now we can write geometric means of ${x_1},{x_2},{x_3},........,{x_n}$
GM$ = {\left( {{x_1}{x_2}{x_3}........{x_n}} \right)^{\dfrac{1}{n}}}$
And we know ${x_1} + {x_2} + {x_3} + ........ + {x_n} = 20$ and ${x_1}{x_2}{x_3}........{x_n} > 1$
And we know AM$ \geqslant $GM
$\dfrac{{{x_1} + {x_2} + {x_3} + ........ + {x_n}}}{n} \geqslant {\left( {{x_1}{x_2}{x_3}........{x_n}} \right)^{\dfrac{1}{n}}}$
As ${x_1} + {x_2} + {x_3} + ........ + {x_n} = 20$
$\dfrac{{20}}{n} \geqslant {\left( {{x_1}{x_2}{x_3}........{x_n}} \right)^{\dfrac{1}{n}}}$
And we know that ${x_1}{x_2}{x_3}........{x_n} > 1$, and now taking ${n^{th}}$ power both side
\[{\left( {\dfrac{{20}}{n}} \right)^n} \geqslant \left( {{x_1}{x_2}{x_3}........{x_n}} \right)\]
And ${x_1}{x_2}{x_3}........{x_n} > 1$, so
\[{\left( {\dfrac{{20}}{n}} \right)^n} \geqslant \left( 1 \right)\]
So $\left( {\dfrac{{20}}{n}} \right)$ must be greater than $1$
$\left( {\dfrac{{20}}{n}} \right) > \left( 1 \right)$. So, $20 > n$
So maximum number of $n = 19$
As $n$ can be integer only.
So the correct answer is Option C.

Note: You can apply AM$ \geqslant $GM anywhere we know if $(a,b)$ are two elements then AM$ = \dfrac{{a + b}}{2}$ and GM$ = \sqrt {ab} $
So always AM$ \geqslant $GM
Equality sign is possible when $a = b$
AM$ = $GM i.e. $a = b$