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What is the remainder of $62831853$ modulo $11$?
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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 $7$.
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Sir, Which formula it is?
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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

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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,

$= (3 + 5 \times 10 + 8 \times 1 + 1 \times 10 + 3 \times 1 + 8 \times 10 + 2 \times 1 + 6 \times 10 ) \mod 11$

$= ((3 + 8 + 3 + 2) + 10(5 + 1 + 8 + 6)) \mod 11$

$= ((3 + 8 + 3 + 2) \mod 11 + 10(5 + 1 + 8 + 6) \mod 11) \mod 11$

 $= ((3 + 8 + 3 + 2) \mod 11 + ((10 \mod 11) \times (5 + 1 + 8 + 6) \mod 11)) \mod 11)  \mod 11$

 $= ((3 + 8 + 3 + 2) \mod 11 + (-1 \times (5 + 1 + 8 + 6) \mod 11) \mod 11) \mod 11$

 $= ((3 + 8 + 3 + 2)  -1 \times (5 + 1 + 8 + 6) ) \mod 11$

 $= ((3 + 8 + 3 + 2)  -(5 + 1 + 8 + 6)) \mod 11$

$= -4 \mod 11$

$= -4+11 = 7$
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