Question

Dau coroana........ ✨ ​

Answer 1

Răspuns:

0

Explicație pas cu pas:

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

$(a + b)|({a}^{n} + {b}^{n}), \: n \: impar$

$a ≡ -b \: (mod \: a + b) => {a}^{n} ≡ {( - b)}^{n} (mod \: a + b)$

${a}^{n} ≡ -{b}^{n} \: (mod \: a + b), \: n \: impar$

$=> {a}^{n} + {b}^{n} ≡ 0 \: (mod \: a + b)$

${1}^{2019} + {29}^{2019} ≡ 0 \: mod \: (1 + 29) ≡ 0 \: mod \: 30$

${2}^{2019} + {28}^{2019} ≡ 0 \: mod \: 30$

...

${14}^{2019} + {16}^{2019} ≡ 0 \: mod \: 30$

${15}^{2019} ≡ 15 \: mod \: 30$

${30}^{2019} ≡ 0 \: mod \: 30$

Astfel:

${1}^{2019} + {2}^{2019} + {3}^{2019} + ... + {30}^{2019} ≡ 15 \: mod \: 30$

2010 = 67×30, deci:

${1}^{2019} + {2}^{2019} + ... + {2010}^{2019} ≡ 15 \: mod \: 30$

iar:

${2011}^{2019} + {2012}^{2019} + ... + {2019}^{2019} ≡ \\ ≡ (1 + 8 + 27 + 4 + 5 + 6 + 13 + 2 + 9) \: mod \: 30 \\ ≡ 75 \: mod \: 30 ≡ 15 \: mod \: 30$

în final:

${1}^{2019} + {2}^{2019} + {3}^{2019} + ... + {2019}^{2019} ≡ 0 \: mod \: 30$