Question

VA ROGG
DAU COROANĂ ​

Answer 1

Răspuns:

ultima cifră

Explicație pas cu pas:

21.

$u( {2018}^{2021} + {2019}^{2020} + {2020}^{2021} ) = u( {8}^{2021}) + u({9}^{2020}) + u({0}^{2021} )$

$= u({8}^{4\cdot505} \cdot 8) + u( {9}^{2\cdot1010} ) + u(0) = u({8}^{4} \cdot 8) + u( {9}^{2} ) + 0$

$= u(6 \cdot 8) + 1 = 8 + 1 = \red{\bf 9}$

restul împărțirii la 10 al numărului este 9

22. a)

$S = {2}^{0} + {2}^{1} + {2}^{2} + ... + {2}^{2019}$

$2S = 2 \cdot ({2}^{0} + {2}^{1} + {2}^{2} + ... + {2}^{2019})$

$2S = {2}^{1} + {2}^{2} + ... + {2}^{2019} + {2}^{2020}$

$2S + 1 = 1 + {2}^{1} + {2}^{2} + ... + {2}^{2019} + {2}^{2020}$

$2S + 1 = ({2}^{0} + {2}^{1} + {2}^{2} + ... + {2}^{2019}) + {2}^{2020}$

$2S + 1 = S + {2}^{2020} \implies \red {\bf S + 1 = {2}^{2020}}$

22. b)

$u( {2}^{2020} ) = u( {2}^{4\cdot505}) = u( {2}^{4}) = \red{\bf 6}$

23.

$a = 1 + {3}^{1} + {3}^{2} + ... + {3}^{2019}$

$3a = 3 \cdot (1 + {3}^{1} + {3}^{2} + ... + {3}^{2019})$

$3a = {3}^{1} + {3}^{2} + ... + {3}^{2019} + {3}^{2020}$

$3a + 1 = 1 + {3}^{1} + {3}^{2} + ... + {3}^{2019} + {3}^{2020}$

$3a + 1 = (1 + {3}^{1} + {3}^{2} + ... + {3}^{2019}) + {3}^{2020}$

$3a + 1 = S + {3}^{2020}$

$2a + 1 = {3}^{2020} = {3}^{2 \cdot 1010} = ({3}^{1010})^{2}$

$\implies \red {\bf 2a + 1 = ({3}^{1010})^{2}}$

Answer 2

Răspuns:

Explicație pas cu pas:

21.

U(8^1) = 8

U(8^2) = 4

U(8^3) = 2

U(8^4) = 6

U(8^5) = 8

ultima cifra se repeta din 4 in 4

2021 : 4 = 505 rest 1

U(8^2021) = 8

U(9^1) = 9

U(9^2) = 1

U(9^3) = 9

ultima cifra se repeta din 2 in 2

U(9^2020) = 1

0 la orice putere este 0

U(2020^2019) = 0

U(2018^2021 + 2019^2020 + 2020^2019) = U(8 + 1 + 0) = 9

restul impartirii la 10 a unui numar care se termina in 9 este 9

________________

22.

S = 2^0 + 2^1 + 2^2 + ... + 2^2019

2S = 2^1 + 2^2 + 2^3 + ... + 2^2020

2S - S = 2^1 + 2^2 + 2^3 + ... + 2^2020 - 2^0 - 2^1 - 2^2 - ... - 2^2019

S = 2^2020 - 2^0 = 2^2020 - 1

S + 1 = 2^2020 - 1 + 1 = 2^2020

U(2^1) = 2

U(2^2) = 4

U(2^3) = 8

U(2^4) = 6

U(2^5) = 2

ultima cifra se repeta din 4 in 4

2020 : 4 = 505 rest 0

U(2^2020) = 6

_________________

23.

a = 1  + 3 + 3^2 + 3^3 + ... + 3^2019

3a = 3 + 3^2 + 3^3 + 3^4 + ... + 3^2020

3a - a = 3 + 3^2 + 3^3 + 3^4 + ... + 3^2020 - 1  - 3 - 3^2 - 3^3 - ... - 3^2019

2a = 3^2020 - 1

2a + 1 = 3^2020 = (3^1010)^2 = patrat perfect