Question

Dau coroana!! Am nevoie derezolvare!​

Answer 1

Răspuns:

21, 22, 23

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 a numărului este 9

22.a)

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

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

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

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

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

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

b)

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

23.

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

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

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

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

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

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

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

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