2 votes 2 votes closed as a duplicate of: ISRO2008-22 The number of two input multiplexers required to construct a $2^{10}$ input multiplexer is, A. $31$ B. $10$ C. $127$ D. $1023$ Digital Logic digital-logic test-series multiplexer + – Pradip Nichite asked Jan 24, 2016 • closed Oct 27, 2023 by Hira Thakur Pradip Nichite 2.8k views comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 5 votes 5 votes No of $1^{st}$ level $2:1$MUX $= \frac{2^{10}}{2} = 2^9$ No of $2^{nd}$ level $2:1$MUX $= \frac{2^9}{2} = 2^8$ . . . no of $10^{th}$ level $2:1$MUX $= \frac{2^1}{2} = 1$ sum of all, Number of $2:1$MUX required $= 2^9 + 2^8 + ......... + 1 = 1023$ Digvijay Pandey answered Jan 24, 2016 • selected Jan 24, 2016 by Praveen Saini Digvijay Pandey comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Praveen Saini commented Sep 23, 2016 reply Follow Share mx1 mux using nx1 mux 0 votes 0 votes GateAspirant999 commented Sep 23, 2016 reply Follow Share Can you please put values of m and n in the formula ceil(m-1/n-1) to get 1023 so that I can get it. 0 votes 0 votes Praveen Saini commented Sep 24, 2016 reply Follow Share No. best way to follow as given in above solution. 0 votes 0 votes Please log in or register to add a comment.