Last Digit Of $n^n$ Summation

If you don't mind, I'll explain.

You see, $a \equiv b \pmod m \implies a^n \equiv b^n \pmod m$ So, if you consider only the unit's digits, then you are essentially taking modulo 10 at every step. So, say, ... ${13}^n \equiv 3^n \pmod {10}$ because $13 \equiv 3 \pmod {10}$ . Hence, in the summation in question, the base repeats every after every ten numbers.

As for the power, they have a periodicity or cyclicity in modulo 10. For numbers ending in $0, 1, 5, 6$ the cycle is of length 1 (Last digit of any positive power of such a number is same as that of the number itself like, ${11}^2 = 121$ has last digit 1 which is same as 11).

For those ending in $2, 3, 7, 8$ , the cycle is of length 4. (For these numbers, $a^{4k+n} \equiv a^n \pmod {10}$ where a is one of these numbers)

For those ending in 4 and 9, the cycle is of length 2.

Hence, the lowest common multiple of these powers is 4. Thus, for all numbers $a^{4k+n} \equiv a^n \pmod {10}$ where $n$ is positive.

So powers repeat every four terms and bases repeat every 10 terms. Hence doing the LCM again, $\text{base}^{\text{powers}}$ repeats after every 20 terms atmost in integers modulo 10.

That is, $(a+10j)^{4k+n} \equiv a^n \pmod {10}$ So, $(n+20k)^{n+20k} \equiv n^n \pmod {10}$

Note : So we can group the terms as such: $\dfrac{1000}{20} \left(1^1 + 2^2 + \cdots + 20^{20} \right) \equiv \boxed0 \pmod{10}$ .

You should cross check manually (use a calculator or something ... ) to see whether this holds. Hope this helps!

Let $N$ be the given number. Then:

$\begin{aligned} N & \equiv \left(\sum_{k=1}^{1000} k^k \right) \text{ (mod 10)} & \small \color{#3D99F6} \text{Let }k=10i + j \text{ as follows.} \\ & \equiv \left(\sum_{i=0}^{99} \sum_{j=0}^9 (10i+j)^{10i+j} +1000^{1000} \right) \text{ (mod 10)} & \small \color{#3D99F6} j \ne 0 \text{ when }i = 0 \\ & \equiv \left(\sum_{i=0}^{99} \sum_{j=0}^9 j^{10i+j} \right) \text{ (mod 10)} \\ & \equiv \left(\sum_{j=0}^9 j^j \sum_{i=0}^{99} j^{10i} \right) \text{ (mod 10)} \\ & \equiv \left(\sum_{j=0}^9 a_j \right) \text{ (mod 10)} & \small \color{#3D99F6} \text{where } a_j = j^2 \sum_{i=0}^{99} j^{10i} \end{aligned}$

Then we have

• $j=0 \implies a_0 \equiv 0 \text{ (mod 10)}$ .
• $j=1 \implies a_1 \equiv 1^1 \displaystyle \sum_{i=0}^{99} 1 \equiv 100 \equiv 0 \text{ (mod 10)}$ .
• $j= 2 \implies \displaystyle a_2 \equiv 2^2 \sum_{i=0}^{99} 2^{10i} \equiv 4 \sum_{i=0}^{99} 4^{5i} \equiv \color{#3D99F6} \sum_{i=1}^{100} 4^{5i} \equiv 4\times 50(4+6) \equiv 0 \text{ (mod 10)}$ . Note that $\color{#3D99F6} 4^n \equiv \begin{cases} \text{4 (mod 10)} & \small \text{when }n \text{ is odd.} \\ \text{6 (mod 10)} & \small \text{when }n \text{ is even.} \end{cases}$
• $j=3 \implies \displaystyle a_3 \equiv 3^3 \sum_{i=0}^{99} 3^{10i} \equiv 7 \sum_{i=0}^{99} 9^{5i} \equiv 7\sum_{i=0}^{99} (10-1)^{5i} \equiv 7\color{#3D99F6} \sum_{i=0}^{99} (-1)^{5i} \equiv 0 \text{ (mod 10)}$ . Note that $\color{#3D99F6} (-1)^n \equiv \begin{cases} \text{-1 (mod 10)} & \small \text{when }n \text{ is odd.} \\ \text{1 (mod 10)} & \small \text{when }n \text{ is even.} \end{cases}$
• $j= 4 \implies \displaystyle a_4 \equiv 4^4 \sum_{i=0}^{99} 4^{10i} \equiv 6 \color{#3D99F6} \sum_{i=0}^{99} 6 \equiv 6(100\times 6) \equiv 0 \text{ (mod 10)}$ . Note that $\color{#3D99F6} 4^n \equiv 6 \text{ (mod 10)}$ , when $n$ is a positive even integer.
• $j= 5 \implies \displaystyle a_5 \equiv 5^5 \sum_{i=0}^{99} 5^{10i} \equiv 5 \color{#3D99F6} \sum_{i=0}^{99} 5 \equiv 5(100\times 5) \equiv 0 \text{ (mod 10)}$ . Note that $\color{#3D99F6} 5^n \equiv 5 \text{ (mod 10)}$ for all $n \in \mathbb N$ .
• $j= 6 \implies \displaystyle a_6 \equiv 6^6 \sum_{i=0}^{99} 6^{10i} \equiv 6 \color{#3D99F6} \sum_{i=0}^{99} 6 \equiv 6(100\times 6) \equiv 0 \text{ (mod 10)}$ . Note that $\color{#3D99F6} 6^n \equiv 6 \text{ (mod 10)}$ for all $n \in \mathbb N$ .
• $j=7 \implies \displaystyle a_7 \equiv 7^7 \sum_{i=0}^{99} 7^{10i} \equiv 7^7 \sum_{i=0}^{99} 49^{5i} \equiv 7^7 \sum_{i=0}^{99} 9^{5i} \equiv 7^7\sum_{i=0}^{99} (10-1)^{5i} \equiv 7^7 \sum_{i=0}^{99} (-1)^{5i} \equiv 0 \text{ (mod 10)}$ .
• $j= 8 \implies \displaystyle a_8 \equiv 8^8 \sum_{i=0}^{99} 8^{10i} \equiv 8^8 \sum_{i=0}^{99} 4^{15i} \equiv 8^8 \times 50(4+6) \equiv 0 \text{ (mod 10)}$ .
• $j=9 \implies \displaystyle a_9 \equiv 9^9 \sum_{i=0}^{99} 9^{10i} \equiv 9^9 \sum_{i=0}^{99} (10-1)^{10i} \equiv 9^9 \sum_{i=0}^{99} 1 \equiv 9^9 \times 100 \equiv 0 \text{ (mod 10)}$ .

We note that $a_0 \equiv a_1 \equiv a_2 \equiv \cdots \equiv a_9 \equiv 0 \text{ (mod 10)}$ , therefore, $N \equiv \displaystyle \sum_{j=0}^0 a_j \equiv \boxed{0} \text{ (mod 10)}$ .

Mathematica

Mod[Total@Array[#^# &, 1000], 10]

returns 0

Very Nice problem.