2^2016-1 is (2^4)^504-1 or 0 and we are done

One way of solving this is to show that the units digit is either 5 or 0, in which case it must be divisible by 5 Now, 2017^2016≡7^2016≡7^(4×504)≡7^4≡1 (mod 10) 2016^2017≡6^2017≡6 (mod 10) Subtracting, we obtain 2017^2016 - 2016^2017≡1 - 6≡-5≡5 (mod 10)

As Vikram pointed out, we can just examine the last digit of the number. For a number to be divisible by $5$, the last digit of that number must be either $5$ or $0$. So we take this problem $\pmod {10}$. First, let's examine $2017$. We have that $2017 \equiv 7 \pmod{10}$. We also have that $2016 \equiv 6 \pmod{10}$. Rewriting what we have: $7^{2016} - 6^{2017} \pmod{10}$. So really what we have is $1 - 6 \pmod{10}$, or $-5 \pmod{10}$ or $5 \pmod{10}$. We see that the last digit of this should be $5$, meaning it is divisible by $5$. Done.

Elaborating on claserken's post, $2016$ is $1 \text{mod} 5$, and any power of one must be one, so the second term is congruent to $1 \text{mod} 5$. For the second term, we use Euler's totient theorem. We get that $n^{\phi{5}}=1 \text{mod} 5$, or $n^4=1 \text{mod} 5$ for any positive integer $n$. Thus, we can put the first term in the $(2^4)^{504}=1 \text{mod} 5$. If we subtract the two terms, we get $0 \text{mod} 5$, which means that the expression must be divisible by $5$.

$2017^{4}{\equiv}1 (mod 5)$ so $2017^{2016}{\equiv}1 (mod 5)$ now $2016{\equiv} 1 (mod 5)$ so $2016^{2017}{\equiv}1 (mod 5)$ so $2017^{2016}-2016^{2017}{\equiv} 0 (mod 5)$....solved

Simple...fermats little theorem

very very strange problem for mational level

\[2017^{2016}-2016^{2017} \equiv 2^{2016}-1 \equiv (-1)^{1008}-1 \equiv 0 \pmod{5}\]

Simple solution: The last digit of the ${{2017}^{2016}}$ it is $1$, and the last digit of ${{2016}^{2017}}$ it is $6$. So the difference have last digit $5$.

dangerousliri wrote: Show that the number $2017^{2016}-2016^{2017}$ is divisible by $5$ . We take this mod 5. $2017^{2016}-2016^{2017}\equiv2^{2016}-1^{2017}\pmod5$ Obviously, $2017$ is prime, so $2^{2016}\equiv1\pmod5$ by FLT. Thus: $2017^{2016}-2016^{2017}\equiv1-1\equiv0\pmod5$

2017≡2 mod 5 2017²⁰¹⁶≡2²⁰¹⁶ mod 5 2²≡-1mod 5 2²⁰¹⁶≡1mod 5 So,2017²⁰¹⁶≡1 mod 5 2016≡1 mod 5 2016²⁰¹⁷≡1 mod 5 So, 2017²⁰¹⁶-2016²⁰¹⁷≡0 mod 5 #@Krishijivi

First, we find the remainder when $2017^{2016}$ is divided by $5$: $2017^4 \equiv 1 \pmod{5} \Rightarrow 2017^{2016} \equiv 1 \pmod{5}$ Then the same thing for $2016^{2017}$: $2016 \equiv 1 \pmod{5} \Rightarrow 2016^{2017} \equiv 1 \pmod{5}$ $2017^{2016}-2016^{2017} \equiv 1-1 \equiv 0 \pmod{5}$ $\blacksquare$