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.
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Sebastien Traglia
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3 years, 11 months ago

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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Tan Li Xuan
·
3 years, 11 months ago

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@Tan Li Xuan
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Did you learn this from Mathematical Circle? Just curious cause i did so :P
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Soham Chanda
·
3 years, 11 months ago

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@Soham Chanda
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No,I didn't. I learned it from my olympic maths books :D
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Tan Li Xuan
·
3 years, 11 months ago

@Superman Son
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Are you asking what is olympic maths?Olympic Maths is basically a harder version of maths,and uses other methods to solve problems.
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Tan Li Xuan
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3 years, 11 months ago

@Priyansh Sangule
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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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Soham Chanda
·
3 years, 11 months ago

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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. – Sebastien Traglia · 3 years, 11 months ago

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– Alan Liang · 3 years, 11 months ago

can you please elaborateLog in to reply

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. – Tan Li Xuan · 3 years, 11 months ago

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– Soham Chanda · 3 years, 11 months ago

Did you learn this from Mathematical Circle? Just curious cause i did so :PLog in to reply

– Tan Li Xuan · 3 years, 11 months ago

No,I didn't. I learned it from my olympic maths books :DLog in to reply

– Superman Son · 3 years, 11 months ago

whats that?Log in to reply

– Tan Li Xuan · 3 years, 11 months ago

Are you asking what is olympic maths?Olympic Maths is basically a harder version of maths,and uses other methods to solve problems.Log in to reply

– Priyansh Sangule · 3 years, 11 months ago

@Soham What is mathematical circle ?Log in to reply

– Zi Song Yeoh · 3 years, 11 months ago

A book.Log in to reply

– Soham Chanda · 3 years, 11 months ago

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. :DLog in to reply