what is the remainder when $7^{2015}$ is divided by $25$? The question says that what is the remainder when $7^{2015}$ is divided by $25$. Please help me out, I am out of ideas (I know that I have to find out the last two digits and I will be done... but I only know how to find the units digit.) and I need your help.
 A: $7^2\equiv -1\pmod{25}$, so $ 7^4\equiv 1\pmod{25}$, so $7^{2015}\equiv 7^{2015\bmod 4}\equiv 7^3\equiv -7\pmod{25}$.
A: $$7^2\equiv-1\pmod{25}$$
As $2015=2\cdot1007+1,$
$$7^{2015}=7\cdot(7^2)^{1007}\equiv7\cdot(-1)^{1007}\pmod{25}\equiv-7\equiv-7+25$$
A: In order to find the last two digits, you need to compute $7^{2015}\pmod{100}$.
According to Euler's theorem, if $\gcd(a,n)=1$, then $a^{\phi(n)}\equiv1\pmod{n}$:


*

*$\gcd(7,100)=1\implies7^{\phi(100)}\equiv\color\red{7^{40}}\equiv1\pmod{100}\implies$

*$7^{2015}\equiv7^{40\cdot50+15}\equiv(\color\red{7^{40}})^{50}\cdot7^{15}\equiv1^{50}\cdot7^{15}\equiv4747561509943\equiv43\pmod{100}\implies$

*$7^{2015}\bmod{25}=43\bmod{25}=18$

But please note that you might as well apply this theorem directly on $25$.
According to Euler's theorem, if $\gcd(a,n)=1$, then $a^{\phi(n)}\equiv1\pmod{n}$:


*

*$\gcd(7,25)=1\implies7^{\phi(25)}\equiv\color\red{7^{20}}\equiv1\pmod{25}\implies$

*$7^{2015}\equiv7^{20\cdot100+15}\equiv(\color\red{7^{20}})^{100}\cdot7^{15}\equiv1^{100}\cdot7^{15}\equiv4747561509943\equiv43\pmod{25}\implies$

*$7^{2015}\bmod{25}=43\bmod{25}=18$
A: The answer could be solved using cyclicity.
$7^1=7$ 
$7^2=49$ 
$7^3=343$
$7^4=2401.$In each last digit is repeating after 4 powers like
$7^5=16807$ 
$7^6=117649$ 
$7^7=823543$ 
$7^8=5764801.$ The last digits 7,9,3,1 are repeating every 4powers.  So $7^{2015}/25$ what will be remainder. We have to divide 2015 by 4 as cyclicity of 7 is 4. 2015/4 remainder will be 3. In 7’s cyclicity 7^3=343 so we have to divide 343 by 25 and check the remainder. The remainder is 18 hence it is the answer.
Therefore when $7^{2015}$ is divided by $25$ the remainder is $18$. Such an easy way.
A: I imagine you found the last digit by looking for a pattern when you do powers of 7 like this:
$7^0=1$
$7^1=7$
$7^2=49\equiv9\pmod{10}$
$7^3\equiv9\times7\equiv63\equiv3\pmod{10}$
$7^4\equiv3\times7\equiv21\equiv1\pmod{10}$
So every 4 powers brings you back to 1. So:
$$7^{2015}=7^{4\times503+3}\equiv7^3\equiv3\pmod{10}$$
You can do the same by looking at the remainder each step when you divide by 25.
$7^0=1$
$7^1=7$
$7^2=49\equiv-1\pmod{25}$
$7^3=-1\times7\equiv-7\pmod{25}$
$7^4\equiv-7\times7\equiv-49\equiv1\pmod{25}$
So it follows the same pattern so it has similar working:
$$7^{2015}=7^{4\times503+3}\equiv7^3\equiv-7\equiv18\pmod{25}$$
A: $7$ has order $4$ modulo 25, hence $\;7^{2015}=7^{2015\bmod4}=7^{-1}\mod25$. Now Bézout's identity: $\;2\cdot 25-7\cdot 7=1\;$  shows  $7^{-1}\equiv -7\mod 25$.
A: $\varphi(25)=20$ implies $7^{2015}\equiv 7^{15} \equiv 7\cdot(49)^7 \equiv -7\pmod{25}$.
A: You can find the last two digits just like you find the last digit. 
{7^0} = 01
{7^1} = 07
{7^2} = 49
(7^3} = ...43
{7^4} = ...01
so that's where the sequence repeats. {7^2015} ends in 43 because 2015 = 4k + 3. So {7^2015} = 18 (modulo 25). 
If you were not looking for the remainder modulo 25 (which is easy to see because 25 divides 100), but say 35, you would calculate {7^k} modulo 35 until the values repeat. 
For a larger number, say modulo 5100, you'd calculate the values modulo 3, 17, and 100, and combine them. 
