Question

va rog aa ma ajutați ​

Answer 1

Răspuns:

Explicație pas cu pas:

c)

U(4^n) ∈ {4,6}, ∀ n ≥ 1, deci dublete

2019:2 = 1009 si rest 1 deci

U(4^2019) = 4, prima cifra din dublet, dupa cum indica restul impartirii.

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In mod analog cercetam 7^2020,

U(7^n) ∈ {1, 9, 3}

2020 : 3 = 673 rest 1

U(7^2020) = 1, primul element al multimii terminatiilor, dupa ce-am luat in consideratie 673 de triplete.

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U(8^n) ∈ {8, 4, 2, 6}

2021:4 = 505 si rest 1

U(82021) = 8, primul element al mutimii terminatiilor, dupa cele 505 de triplete.

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U(9^n) ∈ {9, 1}

2022:2=1011

U(9^2022) = 1

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4^2019 + 7^2020 + 8^2021 + 9^2022 = U(4+1+8+1) =

U(14) =

4.

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a)

U(3^n) ∈ {3, 9, 7, 1}

2019:4 = 504 rest 3 deci U(3^2019) = 7

U(5*n) ∈ {5}, deci U(5^2020) = 5

1^2021 = 1

Ultima cifra a sumei date este

U(7+5+1) = U(13) = 3.

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b)

U(2^n) ∈ {2, 4, 8, 6}

2019;4 = 504 rest 3, deci

U(2^2019) = 8, a 3-a cifra din tripetul de terminatii.

6^2020

U(6^n) = 6, deci U(6^2020) = 6

2019^0 = 1 si astfel avem ultima cifra a sumei cerute este

U(8+6+1) = U(15) = 5.