Question

Demonstrati ca a=2^0+2^1+2^2+2^3+...+2^2019 este divizibil cu 5

Answer 1

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

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

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$a = {2}^{2020} - {2}^{0}$

$a = {2}^{2020} - 1$

$u(a) = u( {2}^{2020} ) - u(1)$

$u( {2}^{2020} ) = ?$

2¹=2. 4k=u(...6).

2²= 4. 4k+1= (...2)

2³=8. 4k+2= (...4)

2⁴=16. 4k+3= (...8)

2020:4= 502 rest 0=> 4k= (...6)

$= > u( {2}^{2020} ) = 6$

$u(a) = 6 - 1 = 5$

u(a)=5 => numarul a se divide cu 5