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Comments
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 31 to 321. 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 32011 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
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
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 31 to 321. 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 32011 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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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
A book.
No,I didn't. I learned it from my olympic maths books :D
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whats that?
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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.