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Math Help - Please help me with these problems.

  1. #1
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    Please help me with these problems.

    1)Prove that if n is an odd positive integer, then
    N = 2269n + 1779n + 1730n 1776n
    is an integer multiple of 2001... 2)Factor the expression
    30(a2 + b2 + c2 + d2) + 68ab 75ac 156ad 61bc 100bd + 87cd: 3) Prove that for every x∈R is true this equation: x^8+x^6-x^3-x+1>0,,, 4) Make the add:1^2+2^2-3^2+… +〖( n-1)〗^2+n^2 which n∈ N ,,,,5) Find the last digit of 777^777
    Last edited by blertta; February 8th 2008 at 06:10 AM. Reason: Im sorry for my english....
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  2. #2
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    Quote Originally Posted by blertta View Post

    1)Prove that if n is an odd positive integer, then
    N = 2269n + 1779n + 1730n 1776n
    is an integer multiple of 2001...

    2)Factor the expression
    30(a2 + b2 + c2 + d2) + 68ab 75ac 156ad 61bc 100bd + 87cd:

    3) Prove that for every x∈R is true this equation: x^8+x^6-x^3-x+1>0,,,

    4) find this sum :1^2+2^2-3^2+… +〖( n-1)〗^2+n^2 which n∈ N ,,,,

    5) Find the last digit of 777^777

    1) Please correct the typo

    2) What does i represent here?
    4) What's happening with the signs? Are they + + + + or + - + - ?

    5) It's useful to know something (interesting?) about integers of the form a^{4n+1}
    This is all you need if you don't have to prove it.
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  3. #3
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    Lexington, MA (USA)
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    Hello, blertta!

    Some of the problems are impossible to read.


    4) Find this sum: . 1^2+2^2-3^2+ \hdots + n^2 where  n \in N
    There is a formula for this series: . S_n \;=\;\frac{n(n+1)(2n+1)}{6}

    But I suppose they want to see its derivation . . .



    5) Find the last digit of 777^{777}
    We are concerned with the last digit (only) of any number.

    Note the endings of consecutive powers of 7.

    . . \begin{array}{ccc} 7^1 & \to & 7 \\ 7^2 & \to & 9 \\ 7^3 & \to & 3 \\ 7^4 & \to & 1 \\ 7^5 & \to & 7 \\ \vdots & & \vdots \end{array}

    We see that 7^4 ends in 1 ... and the sequence repeats.


    We have: . 7^{777} \;=\;7^{776}\cdot 7 \;=\;(7^4)^{194}\cdot 7 \;\to \;1^{194}\cdot 7 \;=\;7

    Therefore, the last digit of 777^{777} is 7.

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