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I’m trying to know how to calculate step by step the next equation applying the module in each operation:

$2^\left(4 \times \frac{6}{8}\right) \pmod{11}$

I know that if I just solve the whole equation and then I apply mod 11, the result is 8.

$4 \times \frac{6}{8}= 3$

$2^3 = 8$

$8 \pmod {11} = 8$

But if I try to do with modular arithmetic is not working for me:

$8^{-1} \pmod{11} = 7$

$6 \times 7 \pmod{11} = 9$

$4 \times 9 \pmod{11} = 3$

$2 ^ 3 \pmod{11} = 9$

Something similar happened when I tried to solve first what is inside the brackets in an expression like this:

$2^\left(4\times 6 \times 8\right) \pmod{11} $

But I changed the way to solve the equation and it works

$\left(\left(2^4\right) ^6\right)^8 \pmod{11} $

Now with the division I have no idea what to do, like in the first example I showed.

Any idea how to solve it?

Jolo
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  • You're not supposed to take mod 11 of any part of the exponent, because the exponent is just "how many times it's multiplied by itself", and not really a number that is congruent to anything. Instead, what you can do, is exactly what you did in the last bit, to factor the exponent and then take mod 11 of the different parts. – Matti P. Sep 14 '18 at 10:41
  • can you show me how to factorise $2^\left(4 \times \frac{6}{8}\right) \pmod{11}$ I have no idea what to do with the division – Jolo Sep 18 '18 at 11:23

1 Answers1

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Your modular computation with exponents is meaningless since, by Euler's formula, $$ a^r\equiv a^{r\bmod\varphi(n)}\mod n, $$ so you should compute the exponents modulo $\varphi(11)=10 $. Unfortunately, $8$ is not a unit module $10$.

Bernard
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  • Thanks for your answer @Bernard. I’m in a situation when I have to calculate something like this $$2^\left(4 \times \frac{6}{8}\right) \pmod{11}$$, but if I cannot because 8 is not a unit module 10, what else can I do? – Jolo Sep 18 '18 at 11:15
  • @Jolo: A priori, the exponent is given as an integer. So you first calculate this integer, then reduce it modulo $10$ if necessary. What else can be done? – Bernard Sep 18 '18 at 11:36
  • if I do the inverse of 8 mod 10 I have an error and if I do the inverse of 8 mod 11 I get a wrong value. – Jolo Sep 18 '18 at 11:43
  • As I said, you fist compute the value of $4\times\frac68=3$. Then you have no problem. – Bernard Sep 18 '18 at 11:55
  • can I formulate my question in a different way to you? link – Jolo Sep 19 '18 at 07:52