Question

How many numbers less than \[{2^{34}}\] are coprime to it?
A.\[{2^{12}}\]
B.\[{2^{23}}\]
C.\[{2^{22}}\]
D.None of these

Answer 1

Hint: Here the concept of coprime should be known. Coprime numbers are the numbers with only 1 as their common factor. Also the formula to find the numbers less than a given number can be used. So we will note the formula below.
Formula used:
If the number can be expressed in the form \[N = {a^x}\] then the numbers less than the given number N are given by,
\[N\left( {1 - \dfrac{1}{a}} \right)\].

Complete step-by-step answer:
Now the number given to us is \[{2^{34}}\].
So this is of the form, \[{a^x}\]
So we will put N as \[{2^{34}}\] since the number given is in the form \[{a^x}\] already.
Now using the formula as,
\[N\left( {1 - \dfrac{1}{a}} \right)\]
Putting the values as,
\[ = {2^{34}}\left( {1 - \dfrac{1}{2}} \right)\]
On solving the bracket,
\[ = {2^{34}} \times \dfrac{1}{2}\]
We know that 2 can be written as \[{2^1}\]
So the above expression becomes,
\[ = {2^{34 - 1}}\]
Thus the power changes to,
\[ = {2^{33}}\]
Thus these are the numbers that are coprime but less than the given number. But unfortunately this is not available in the options given.
So option D is the correct answer.
So, the correct answer is “Option D”.

Note: Note that, any number can be expressed in the form of prime number with some power as ,
For example 42 can be written as \[42 = {2^1} \times {3^1} \times {7^1}\]
Now this can be written in formula form as \[N = {a^x}{b^y}{c^z}\]
Thus its coprime numbers less than the given numbers will be,
\[N\left( {1 - \dfrac{1}{a}} \right)\left( {1 - \dfrac{1}{b}} \right)\left( {1 - \dfrac{1}{c}} \right)\]
This is explained because in the problem above we are already given the number in power form. If we come across the questions in number form so first convert it in power form and then solve.