Yesterday I encountered this problem while I was solving a problem.But don't worry the problem has been solved.Could somebody give a solution to this:
$2^{2015}\equiv a(mod{20}).$You have to find 'a'.

@Adarsh Kumar
–
Are you joking? (Really, coz u've solved my problem...)..ANyway...$\phi(20)=8.$ thus, we have that $2^{2015} \pmod {20} = 2^7 \pmod {20}$. Which is nothing but 8!!

2^2015 is a multiple of 4
And 2^2015 leaves a remainder of 3 modulo 5 ( Fermat's little theorem) so, on dividing by 20, I.e.4*5, remainder is 8 as krishna said

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest@Satvik Golechha @Krishna Ar ?

Log in to reply

Answer's $8$

Log in to reply

Yes,I actually know the answer but could you please provide a step-by-step solution?

Log in to reply

$\phi(20)=8.$ thus, we have that $2^{2015} \pmod {20} = 2^7 \pmod {20}$. Which is nothing but 8!!

Are you joking? (Really, coz u've solved my problem...)..ANyway...Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

And it was this problem right?

Log in to reply

Yes,it was this problem but I figured out that the answer would not be 10 so I solved that problem but, I do not know how to find the remainder.

Log in to reply

@Calvin Lin ?

Log in to reply

2^2015 is a multiple of 4 And 2^2015 leaves a remainder of 3 modulo 5 ( Fermat's little theorem) so, on dividing by 20, I.e.4*5, remainder is 8 as krishna said

Log in to reply

Good. Nice solution using CRT. Adarsh Kumar

Log in to reply

yes awesome!!

Log in to reply