Question

An

n -digit binary number is multiplied with another m-digit binary number. What is the

maximum value of the result of the multiplication?​

Answer 1

Given :  n -digit binary number is multiplied with another m-digit binary number.

To find : maximum digits in result of the multiplication

Solution:

n Digit Binary number maximum value  = 2ⁿ - 1

m Digit Binary number maximum value  = 2^m - 1

Multiplication = (2ⁿ - 1)( 2^m - 1)

= 2^(n+m) - 2ⁿ - 2^m + 1  

2ⁿ  and 2^m > 1  

Hence Multiplication < 2^(n+m) - 1

the  maximum value of the result of the multiplication  = n + m digits

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Answer 2

Given:

n -digit binary number is multiplied with another m-digit binary number.

To Find:

Maximum value of the result of the multiplication?​

Solution:

Here, a n-digit binary number and m-digit binary number is given .

Since, we know that,

The largest number with n-digit binary number is 2ⁿ - 1.

Similarly, for m-digit binary number is $2^{m} - 1$.

therefore, when n -digit binary number is multiplied with another m-digit binary number is;

$=(2^{n}-1 )(2^{m}-1 )\\=2^{m+n} -(2^{n}+2^{m} )+1$

Hence, maximum value of the result of the multiplication is $2^{m+n}-(2^{n} +2^{m} ) +1$  .