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claudia_cla2011
user (0)
Pe: 2013-01-10T15:20:08+02:00 In: In:

Problema clasa a 5 a

Aflati ultima cifra a numerelor:
a) 6 la put.23 + 0 la put 23 + 23 la put 1
b) 2 la put 42
c) 2 la put 50 + 3 la put 50
d) 2012 la put 40 + 2011 la put 30 + 2007 la put 20
e) 39 la put 81 + 93 la put 18 + 15 la put 42

P.S. Am scris prescurtat puterea cu put

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7 raspunsuri

  1. gunty
    maestru (V)

    a)

        \[ \begin{array}{l} 6^{23} + 0^{23} + 23^1 = 6^{23} + 23 \\ n......u(n) \\ 6^1 .....6 \\ 6^2 .....6 \\ 6^3 .....6 \\ 6^4 .....6 \\ \vdots \\ \end{array} \]

    Observi, ca ultima cifra a numarului 6^n (unde n e un nr. natural nenul) este 6. Deci u(6^23)=6. E clar ca u(23)=3, deci u(6^23+23)=6+3=9.

    * Incearca sa faci la fel si la celelalte… daca te blochezi eventual, astept intrebari.

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  2. claudia_cla2011
    user (0)

    gunty wrote: a)

        \[ \begin{array}{l} 6^{23} + 0^{23} + 23^1 = 6^{23} + 23 \\ n......u(n) \\ 6^1 .....6 \\ 6^2 .....6 \\ 6^3 .....6 \\ 6^4 .....6 \\ \vdots \\ \end{array} \]

    Observi, ca ultima cifra a numarului 6^n (unde n e un nr. natural nenul) este 6. Deci u(6^23)=6. E clar ca u(23)=3, deci u(6^23+23)=6+3=9.

    * Incearca sa faci la fel si la celelalte… daca te blochezi eventual, astept intrebari.

    si b?

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  4. gunty
    maestru (V)

    Incearca s-o faci tu. Cum am mai spus, daca te blochezi undeva, te ajut.

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  5. claudia_cla2011
    user (0)

    gunty wrote: Incearca s-o faci tu. Cum am mai spus, daca te blochezi undeva, te ajut.

    pai nu prea am idee

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  6. edy8
    maestru (V)

    claudia_cla2011 wrote: Aflati ultima cifra a numerelor:
    a) 6 la put.23 + 0 la put 23 + 23 la put 1
    b) 2 la put 42
    c) 2 la put 50 + 3 la put 50
    d) 2012 la put 40 + 2011 la put 30 + 2007 la put 20
    e) 39 la put 81 + 93 la put 18 + 15 la put 42

    P.S. Am scris prescurtat puterea cu put

    „Prescurtat” se poate scrie :

    6^{23}+0^{23}+23^1

    Acest enunt introdus in TeX apare asa :

    6^{23}+0^{23}+23^1

    Exista si nuante care produc o mai buna percepere a mesajelor transmise

    \it{\Large\bl a) u(6^{23}+0^{23}+23^1) = 6+0+3 = 9

    \it{\Large b) Folosim cateva rezultate generale:\\\;\\u(2^{4k}) = u(2^4)^k = u(16^k) = 6\\\;\\u(2^{4k+1})=2\\\;\\u(2^{4k+2})=4\\\;\\u(2^{4k+3})=8}

    \it{\Large\bl u(2^{42})=u(2^{40+2})=4}

    \it{\Large Alte precizari :\\\;\\u(3^{4k})=u(3^4)^k=u(81^k) = 1\\\;\\u(3^{4k+1})=3\\\;\\u(3^{4k+2})=9\\\;\\u(3^{4k+3})=7}

    \it{\Large\bl c) u(2^{50})=u(2^{48+2})=4\\\;\\ u(3^{50})=u(3^{48+2}) =9\\\;\\u(2^{50}+3^{50})=u(4+9)=3}

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  8. edy8
    maestru (V)

    \it{\Large\bl d) u(2012^{40})=u(2^{40})=u(2^{4\cdot10})=6\\\;\\u(2011^{30})=u(1^{30})=1\\\;\\u(2007^{20})=u(7^{20})=u(7^4)^5=1\\\;\\u(2012^{40}+2011^{30}+2007^{20})=u(6+1+1)=8}

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  9. petrut5
    user (0)

    claudia_cla2011 wrote:
    e) 39 la put 81 + 93 la put 18 + 15 la put 42

    \rm{u(39^{81}+93^{18}+15^{42})=u(9^{81}+3^{18}+5^{42})=\\\;\\=u(9^{80+1}+3^{16+2}+5^{42})=u(9+9+5)=3}

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