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Question: Convert \((146){}_{10}\) to base \(2\) \(A)\) \({(1010010)_2}\) \(B)\) \({(100010010)_2}\) \(C...

Convert (146)10(146){}_{10} to base 22
A)A) (1010010)2{(1010010)_2}
B)B) (100010010)2{(100010010)_2}
C)C) (10010010)2{(10010010)_2}
D)D) (10001010)2{(10001010)_2}

Explanation

Solution

First we have to define what the terms we need to solve the problem are.
The provided problem is based on the binary system. Then a long process is the decimal to binary conversion, usually achieved by splitting the decimal number into two. Numerical values are represented by two different symbols; zero, one knowing the process of division to answer these questions.

Complete step-by-step solution:
According, to convert any given decimal number into its equivalent binary number system.
Binary numbers are representing as base 22, octal numbers are representing as base 88
Decimal numbers are representing as base 1010 and hexadecimal numbers is representing as base 1616
Decimal to binary conversion is a long process which is usually done by dividing the decimal number to 2
Continuous division of integers is carried out until the remainder reaches 0 or 1.
We represent the (146)10(146){}_{10} to base 22, since we know that the binary number is represent by base 2
Let us begin with dividing 1462\dfrac{{146}}{2} so we will get remainder as zero and quotient as 7373
Again, division by 2 we get 732\dfrac{{73}}{2} but in here we get remainder as one and quotient as 3636
Similarly, we approach the same thing further we get 362\dfrac{{36}}{2} quotient as 1818 and remainder zero
And 182\dfrac{{18}}{2} quotient as 99 and remainder zero, 92\dfrac{9}{2} quotient as 44 and remainder one
44\dfrac{4}{4} quotient as 22 and remainder zero and finally 22\dfrac{2}{2} quotient as one and remainder zero
Therefore, writing the remainders in reverse order that is from bottom to top we get (10010010)2{(10010010)_2}
As we clearly see from bottom to top the remainders are 1,0,0,1,0,0,1,01,0,0,1,0,0,1,0 so no other option is possible except option C.

Note: It is possible to solve the above problem using the binary system formula, the numbers can be positioned to the left or the right of the point and its direct implementation in electronic circuits too.