Question

Let f be a function with the following properties:
(i) f(1) = 1

(ii) f(2n) = n.f(n) for any positive integer n.

What is the value of f(2¹⁰⁰) ?

(A) 2⁵⁰⁵⁰

(B) 2⁹⁹

(C) 2¹⁰⁰

(D) 2⁴⁹⁵⁰

Answer with proper explanation.

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: f(1) = 1$

and

$\rm \: f(2n) = nf(n) \: \: \: \forall \: n \: \in \: positive \: integer$

Now, Consider

$\rm \: f(2)$

$\rm \: = \: f(2 \times 1)$

$\rm \: = \: 1 \times f(1)$

$\rm \: = \: 1$

$\rm \: = \: {2}^{0}$

Thus,

$\rm\implies \:\boxed{\sf{ \:\rm \: f(2) = {2}^{0} \: \: }}$

Now, Consider

$\rm \: f( {2}^{2} )$

$\rm \: = \: f(4)$

$\rm \: = \: f(2 \times 2)$

$\rm \: = \: 2f(2)$

$\rm \: = \: 2 \times 1$

$\rm \: = \: 2$

Thus,

$\rm\implies \:\boxed{\sf{ \:\rm \: f( {2}^{2} ) = {2}^{1} \: \: }}$

Now, Consider

$\rm \: f( {2}^{3} )$

$\rm \: = \: f(8)$

$\rm \: = \: f(2 \times 4)$

$\rm \: = \: 4f(4)$

$\rm \: = \: 4 \times 2$

$\rm \: = \: 8$

$\rm \: = \: {2}^{3}$

Thus,

$\rm\implies \:\boxed{\sf{ \:\rm \: f( {2}^{3} ) = {2}^{3} = {2}^{1 + 2} \: \: }}$

Now, Consider

$\rm \: f( {2}^{4} )$

$\rm \: = \: f(16)$

$\rm \: = \: f(2 \times 8)$

$\rm \: = \: 8f(8)$

$\rm \: = \: 8 \times {2}^{1 + 2}$

$\rm \: = \: {2}^{3} \times {2}^{1 + 2}$

$\rm \: = \: {2}^{1 + 2 + 3}$

Thus,

$\rm\implies \:\boxed{\sf{ \:\rm \: f( {2}^{4} ) = {2}^{1 + 2 + 3} \: \: }}$

So, we have by symmetry,

$\begin{gathered}\boxed{\begin{array}{c|c} \bf n & \bf f(2n) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf {2}^{2} & \sf {2}^{1} \\ \\ \sf {2}^{3} & \sf {2}^{1 + 2} \\ \\ \sf {2}^{4} & \sf {2}^{1 + 2 + 3} \end{array}} \\ \end{gathered}$

Now, Consider

$\rm \: f( {2}^{100} )$

can be rewritten as

$\rm \: = \: {2}^{1 + 2 + 3 + - - - + 99}$

We know,

$\boxed{\sf{ \:1 + 2 + 3 + - - - + n = \frac{n(n + 1)}{2} \: \: }}$

So, using this result, we get

$\rm \: = \: {\bigg(2 \bigg) }^{\dfrac{99(99 + 1)}{2} }$

$\rm \: = \: {\bigg(2 \bigg) }^{\dfrac{99(100)}{2} }$

$\rm \: = \: {\bigg(2 \bigg) }^{\dfrac{99(50)}{1} }$

$\rm \: = \: {2}^{4950}$

Hence,

$\rm\implies \:\boxed{\sf{ \:\rm \: f( {2}^{100} ) \: = \: {2}^{4950} \: \: }}$

So, option (D) is correct.