Solution-1

We find the last digit of the number by looking at it modulo \(10\) (finding its remainder when dividing by \(10). Notice that \)2022^{2022} \equiv 2^{2022}(\bmod 10)\(. So we need to find the last digit of \)2^{2022}\(. The sequence \)2^{n}, n=1,2, \ldots\(, is periodic modulo \)10\(. Look at its first terms. * if \)n=1\(, then \)2^{1} \equiv 2 \pmod {10}\(; * if \)n=2\(, then \)2^{2} \equiv 4 \pmod {10}\(; * if \)n=3\(, then \)2^{3} \equiv 8 \pmod {10}\(; * if \)n=4\(, then \)2^{4} \equiv 16 \equiv 6 \pmod {10}\(; * If \)n=5\(, then \)2^{5} \equiv 32 \equiv 2 \pmod {10}\(. This information can be summarized in a table: | \)n\( | \)2^n \pmod {10}\( | | ----- | ---------------- | | \)1\( | \)\mathbf{2}\( | | \)2\( | \)4\( | | \)3\( | \)8\( | | \)4\( | \)6\( | | \)5\( | \)\mathbf{2}\( | We see that the sequence \)2^{n}(\bmod 10)\( is periodic with a period \)4\(. Since \)2022=4 \cdot 505+2\(, the sequence will make \)505\( full periods, and its \)2022\(-th term same as its second term, so it is equal to \)4\(. We conclude that the last digit of \)2022^{2022}\( is \)4$.

Atrisinājums-1

Skaitļa pēdējo ciparu noskaidrosim, apskatot doto skaitli pēc moduļa \(10\). Ievērosim, ka \(2022^{2022} \equiv 2^{2022}(\bmod 10)\). Tātad mums jānoskaidro skaitļa \(2^{2022}\) pēdējais cipars.

Virkne \(2^{n}, n=1,2, \ldots\), ir periodiska pēc moduļa \(10\), apskatīsim šīs virknes pirmos locekļus:

• ja \(n=1\), tad \(2^{1} \equiv 2 \pmod {10}\);
• ja \(n=2\), tad \(2^{2} \equiv 4 \pmod {10}\);
• ja \(n=3\), tad \(2^{3} \equiv 8 \pmod {10}\);
• ja \(n=4\), tad \(2^{4} \equiv 16 \equiv 6 \pmod {10}\);
• ja \(n=5\), tad \(2^{5} \equiv 32 \equiv 2 \pmod {10}\).

Šo informāciju ērti apkopot tabulā:

Redzam, ka virkne \(2^{n}(\bmod 10)\) ir periodiska ar perioda garumu \(4\). Tā kā \(2022=4 \cdot 505+2\), tad virknes 2022.locekļa pēdējais cipars būs tāds pats kā virknes 2.locekļa pēdējais cipars, tātad pēdējais cipars būs \(4\). Līdz ar to esam ieguvuši, ka skaitļa \(2022^{2022}\) pēdējais cipars ir \(4\).