Explicație pas cu pas:
[tex]S = {2}^{2021} + {3}^{2022} + {4}^{2023} = {2}^{4 \times 505 + 1} + {3}^{4 \times 505 + 2} + {4}^{1011 \times 2 + 1} = 2 \times {2}^{4 \times 505} + {3}^{2} \times {3}^{4 \times 505} + 4 \times {4}^{2 \times 1011}[/tex]
știm că:
[tex]{2}^{1} = 2; \: {2}^{2} = 4; \: {2}^{3} = 8; \: {2}^{4} = 16 ; \: {2}^{5} = 32 ...[/tex]
[tex]{3}^{1} = 3; \: {3}^{2} = 9; \: {3}^{3} = 27; \: {3}^{4} = 81 ; \: {3}^{5} = 243... [/tex]
[tex]{4}^{1} = 4; \: {4}^{2} = 16; \: {4}^{3} = 64...[/tex]
deci:
[tex]{2}^{4 \times 505 + 1} = = {2}^{1} = 2 \\{3}^{4 \times 505 + 2} = = {3}^{2} = 9 \\ {4}^{1011 \times 2 + 1} = = {4}^{1} = 4 [/tex]
suma ultimelor cifre este: 2 + 9 + 4 = 15
=> S divizibil cu 5
