2 votes 2 votes What is the remainder of $62831853$ modulo $11$? Quantitative Aptitude goclasses_wq1 numerical-answers goclasses quantitative-aptitude number-system modular-arithmetic remainder-theorem 2-marks + – GO Classes asked May 1, 2022 GO Classes 463 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes This is related to homework problems. $62831853 \; \equiv \;-6+2-8+3-1+8-5+3=-4\;\equiv \; 7 \; \mod \; 11)$ The remainder of $62831853$ modulo $11$ is $7$. GO Classes answered May 1, 2022 edited May 1, 2022 by Lakshman Bhaiya GO Classes comment Share Follow See all 3 Comments See all 3 3 Comments reply raja11sep commented Jul 6, 2022 reply Follow Share Sir, Which formula it is? 0 votes 0 votes Deepak Poonia commented Jul 6, 2022 reply Follow Share Sir, Which formula it is? Watch “Lecture 4 - Modular Arithmetic” in the following course: https://www.goclasses.in/s/store/courses/description/2023-Discrete-Mathematics 0 votes 0 votes ankitgupta.1729 commented Jul 7, 2022 reply Follow Share It seems divisibility by 11 rule is applied but not sure whether it is correctly used or not because I have never used it. Not sure whether it should be $−6+2−8+3−1+8−5+3$ or $+6-2+8-3+1-8+5-3.$ It is easy to find remainder after dividing $62831853$ by $11$ but it might not work if question is something like $333...68 \;times \mod 11$ etc. So, following proof might be helpful. $:= 62831853 \mod 11$ $= (3 + 5 \times 10 + 8 \times 10^2 + 1 \times 10^3 + 3 \times 10^4 + 8 \times 10^5 + 2 \times 10^6 + 6 \times 10^7 ) \mod 11$ Now, $10^{even} \mod 11 = 1$ and $10^{odd} \mod 11 = 10$ (can be proved easily) and so, $= (3 + 5 \times 10 + 8 \times 1 + 1 \times 10 + 3 \times 1 + 8 \times 10 + 2 \times 1 + 6 \times 10 ) \mod 11$ $= ((3 + 8 + 3 + 2) + 10(5 + 1 + 8 + 6)) \mod 11$ $= ((3 + 8 + 3 + 2) \mod 11 + 10(5 + 1 + 8 + 6) \mod 11) \mod 11$ $= ((3 + 8 + 3 + 2) \mod 11 + ((10 \mod 11) \times (5 + 1 + 8 + 6) \mod 11)) \mod 11) \mod 11$ $= ((3 + 8 + 3 + 2) \mod 11 + (-1 \times (5 + 1 + 8 + 6) \mod 11) \mod 11) \mod 11$ $= ((3 + 8 + 3 + 2) -1 \times (5 + 1 + 8 + 6) ) \mod 11$ $= ((3 + 8 + 3 + 2) -(5 + 1 + 8 + 6)) \mod 11$ $= -4 \mod 11$ $= -4+11 = 7$ 3 votes 3 votes Please log in or register to add a comment.