Since you can find tens digit of a number n by the result of n mod 100, you can use Euler's Theorem. Thanks to this theorem you know that 3^40 = 1 (mod 100), since φ(100) is 40. Therefore 3^2000 = 1 (mod 100) and
3^2011 = 3^11 (mod 100). 3^11 is 177147 and so 177147 mod 100 = 47. Now you know that
3^2011 mod 100 = 47, and then 3^2011 ends with 4 and 7. 4 is the tens digit.

Find a pattern.\( 03,09,27,81,43,29,87,61,83,49,47,41,23,69,07,21,63,89,67,01,03........\) These are the last 2 digits for the numbers \( 3^{1} \) to \( 3^{21} \). As you can see,after every 20 numbers the last two digits go back to \( 03 \).Hence \( 2011/20=100 (r) 11 \).Then we find the 11th number in the 20 number set which is 47,so tens digit of \( 3^{2011} \) is 4.

@Priyansh Sangule
–
First..as your replying to me I take it for granted it's directed towards me .So you didn't have to write "@Soham" :D
Anyways itsa legen-waitforit-dary. :D

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TopNewestSince you can find tens digit of a number n by the result of n mod 100, you can use Euler's Theorem. Thanks to this theorem you know that 3^40 = 1 (mod 100), since φ(100) is 40. Therefore 3^2000 = 1 (mod 100) and 3^2011 = 3^11 (mod 100). 3^11 is 177147 and so 177147 mod 100 = 47. Now you know that 3^2011 mod 100 = 47, and then 3^2011 ends with 4 and 7. 4 is the tens digit.

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Find a pattern.\( 03,09,27,81,43,29,87,61,83,49,47,41,23,69,07,21,63,89,67,01,03........\) These are the last 2 digits for the numbers \( 3^{1} \) to \( 3^{21} \). As you can see,after every 20 numbers the last two digits go back to \( 03 \).Hence \( 2011/20=100 (r) 11 \).Then we find the 11th number in the 20 number set which is 47,so tens digit of \( 3^{2011} \) is 4.

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Did you learn this from Mathematical Circle? Just curious cause i did so :P

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@Soham What is mathematical circle ?

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No,I didn't. I learned it from my olympic maths books :D

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