Question

Determinați ultima cifră a numărului:

a=1+2² +3³ +...+2013²⁰¹³

Indicatie: Se grupează termenii sumei câte zece și se arată că ultima cifră a fiecărei grupe este aceeași, zero.

VĂ ROG FRUMOS AJUTAȚI-MĂ!!​

Answer 1

Răspuns:

7

Explicație pas cu pas:

$a = 1 + {2}^{2} + {3}^{3} + {4}^{4} + ... + {2013}^{2013}$

→ grupăm termenii sumei câte 10:

$U(1) + U( {2}^{2} ) + U( {3}^{3} ) + U( {4}^{4} ) + U( {5}^{5} ) + U( {6}^{6} ) + U( {7}^{7} ) + U( {8}^{8} ) + U( {9}^{9} ) + U( {10}^{10} ) = U(1 + 4 + 7 + 6 + 5 + 6 + 3 + 6 + 9 + 0) = U(47) = 7$

$U( {11}^{11} ) + U( {12}^{12} ) + U( {13}^{13} ) + U( {14}^{14} ) + U( {15}^{15} ) + U( {16}^{16} ) + U( {17}^{17} ) + U( {18}^{18} ) + U( {19}^{19} ) + U( {20}^{20} ) = U(1 + 6 + 3 + 6 + 5 + 6 + 7 + 4 + 9 + 0) = U(47) = 7$

→ observăm că ultima cifră a sumei unei grupe de 10 termeni este 7

→ a are 2013-1+1 = 2013 termeni

2013 = 201×10 + 3

→ atunci:

$U(a) = U\Big[U(201 \cdot 7) + U( {2011}^{2011} ) + U( {2012}^{2012} ) + U( {2013}^{2013} )\Big] = U\Big[U(1407) + U( {1}^{2011} ) + U( {2}^{2012} ) + U( {3}^{2013} )\Big] = U(7 + 1 + 6 + 3) = U(17) = \bf 7$

unde:

▪︎ 0 ridicat la orice putere are ultima cifră 0

▪︎ 1 ridicat la orice putere are ultima cifră 1

▪︎ 5 ridicat la orice putere nenulă are ultima cifră 5

▪︎ 6 ridicat la orice putere nenulă are ultima cifră 6

▪︎ ultima cifră a numerelor: 2, 3, 7, 8 ridicate la o putere nenulă se repetă la fiecare 4 puteri consecutive

▪︎ ultima cifră a numerelor: 4 și 9 ridicate la o putere nenulă se repetă la fiecare 2 puteri consecutive

$U(1) = 1 \\ U( {11}^{11} ) = U( {1}^{11} ) = U(1) = 1 \\ U( {21}^{21} ) = U( {1}^{21} ) = U(1) = 1 \\ U( {31}^{31} ) = U( {1}^{31} ) = U(1) = 1$

$U( {2}^{2} ) = 4 \\ U( {12}^{12} ) = U({2}^{12} ) = U( {2}^{4 \cdot 3} ) = U({2}^{4} ) = 6 \\ U( {22}^{22} ) = U({2}^{22} ) = U( {2}^{4 \cdot 5 + 2}) = U({2}^{2} ) = 4 \\ U( {32}^{32} ) = U({2}^{32} ) = U( {2}^{4 \cdot 8}) = U({2}^{4} ) = 6 \\ \implies \\ U( {2}^{4k}) = U({2}^{4} ) = 6 \\ U( {2}^{4k + 2}) = U({2}^{2} ) = 4$

$U( {3}^{3} ) = 7 \\ U( {13}^{13} ) = U({3}^{13} ) = U( {3}^{4 \cdot 2 + 1}) = U({3}^{1} ) = 3 \\ U( {23}^{23} ) = U({3}^{23} ) = U( {3}^{4 \cdot 5 + 3}) = U({3}^{3} ) = 7 \\ U( {33}^{33} ) = U({3}^{33} ) = U( {3}^{4 \cdot 8 + 1}) = U({3}^{1} ) = 3 \\ \implies \\ U( {3}^{4k + 1}) = U({3}^{1} ) = 3 \\ U( {3}^{4k + 3}) = U({3}^{3} ) = 7$

$U( {4}^{4} ) = U( {4}^{2 \cdot 2} ) = U( {4}^{2} ) = 6 \\ U( {14}^{14} ) = U({4}^{14} ) = U( {4}^{2 \cdot 7}) = U({4}^{2} ) = 6 \\ U( {24}^{24} ) = U({4}^{24} ) = U( {4}^{2 \cdot 12}) = U({4}^{2} ) = 6 \\ U( {34}^{34} ) = U({4}^{34} ) = U( {4}^{2 \cdot 17}) = U({4}^{2} ) = 6 \\ \implies \\ U( {4}^{2k}) = U({4}^{2} ) = 6$

$U( {5}^{5} ) = 5 \\ U( {15}^{15} ) = U( {5}^{15} ) = U(5) = 5 \\ U( {25}^{25} ) = U( {5}^{25} ) = U(5) = 5 \\ U( {35}^{35} ) = U( {5}^{35} ) = U(5) = 5$

$U( {6}^{6} ) = 6 \\ U( {16}^{16} ) = U( {6}^{16} ) = U(6) = 6 \\ U( {26}^{26} ) = U( {6}^{26} ) = U(6) = 6 \\ U( {36}^{36} ) = U( {6}^{36} ) = U(6) = 6$

$U( {7}^{7} ) = U( {7}^{4 + 3} ) = U( {7}^{3} ) = 3 \\ U( {17}^{17} ) = U({7}^{17} ) = U( {7}^{4 \cdot 4 + 1}) = U({7}^{1} ) = 7 \\ U( {27}^{27} ) = U( {7}^{27} ) = U( {7}^{4 \cdot 6 + 3} ) = U( {7}^{3} ) = 3 \\ U( {37}^{37} ) = U({7}^{37} ) = U( {7}^{4 \cdot 9 + 1}) = U({7}^{1} ) = 7 \\ \implies \\ U( {7}^{4k + 1}) = U({7}^{1} ) = 7 \\ U( {7}^{4k + 3}) = U({7}^{3} ) = 3$

$U( {8}^{8} ) = U( {8}^{2 \cdot 4} ) = U( {8}^{4} ) = 6 \\ U( {18}^{18} ) = U({8}^{18} ) = U( {8}^{4 \cdot 4 + 2}) = U({8}^{2} ) = 4 \\ U( {28}^{28} ) = U( {8}^{28} ) = U( {8}^{4 \cdot 7} ) = U( {8}^{4} ) = 6 \\ U( {38}^{38} ) = U({8}^{38} ) = U( {8}^{4 \cdot 9 + 2}) = U({8}^{2} ) = 4 \\ \implies \\ U( {8}^{4k}) = U({8}^{4} ) = 6 \\ U( {8}^{4k + 2}) = U({8}^{2} ) = 4$

$U( {9}^{9} ) = U( {9}^{2 \cdot 4 + 1} ) = U( {9}^{1} ) = 9 \\ U( {19}^{19} ) = U({9}^{19} ) = U( {9}^{2 \cdot 9 + 1}) = U({9}^{1} ) = 9 \\ U( {29}^{29} ) = U({9}^{29} ) = U( {9}^{2 \cdot 14 + 1}) = U({9}^{1} ) = 9 \\ U( {39}^{39} ) = U({9}^{39} ) = U( {9}^{2 \cdot 19 + 1}) = U({9}^{1} ) = 9 \\ \implies \\ U( {9}^{2k + 1}) = U({9}^{1} ) = 9$

$U( {10}^{10} ) = U( {0}^{10} ) = 0 \\ U( {20}^{20} ) = U( {0}^{20} ) = U(0) = 0 \\ U( {30}^{30} ) = U( {0}^{30} ) = U(0) = 0 \\U( {40}^{40} ) = U( {0}^{40} ) = U(0) = 0$