Number theory basic

Find the last digit of

1 ¹ + 2 ² + 3 ³ + 4 4 + . . . . . . . . . + 202 0 2020 \large1¹+2²+3³+4^{4}+.........+2020^{2020}


The answer is 4.

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1 solution

Let S S be the given sum. We need to find S m o d 10 S \bmod 10 .

S k = 1 2020 k k (mod 10) k = 1 2020 ( 1 0 k 1 + k ) 1 0 k 1 + k (mod 10) 202 k = 1 10 k k (mod 10) 2 ( 1 1 + 2 2 + 3 3 + 4 4 + 5 5 + 6 6 + 7 7 + 8 8 + 9 9 + 1 0 10 ) (mod 10) 2 ( 1 + 4 + 7 + 6 + 5 + 6 + 3 + 6 + 9 + 0 ) (mod 10) 2 ( 7 ) 4 (mod 10) \begin{aligned} S & \equiv \sum_{k=1}^{2020} k^k \text{ (mod 10)} \\ & \equiv \sum_{k=1}^{2020} \left(10^{k-1}+k\right)^{10^{k-1}+k} \text{ (mod 10)} \\ & \equiv 202 \sum_{k=1}^{10} k^k \text{ (mod 10)} \\ & \equiv 2 \left(1^1+2^2+3^3+4^4+5^5+6^6+7^7+8^8+9^9+10^{10}\right) \text{ (mod 10)} \\ & \equiv 2 \left(1+4+7+6+5+6+3+6+9+0\right) \text{ (mod 10)} \\ & \equiv 2(7) \equiv \boxed 4 \text{ (mod 10)} \end{aligned}

@arifin Ikram , the word "out" after "find" is unnecessary.

Chew-Seong Cheong - 1 year, 8 months ago

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