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Thread: Divisibly Problems

  1. #1
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    Question Divisibly Problems

    Here are 3 problems that i couldn't solve them. I would wonder if anybody help me.

    a) If 5|3a+4b prove that 25|2a^2-3ab+2b^2
    b) If a^9|b^5 prove that a^7|b^4
    c) If a, b, c are natural numbers and a^b|b^a and b^c|c^b, prove that a^c|c^a
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  2. #2
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    Quote Originally Posted by havaliza View Post
    Here are 3 problems that i couldn't solve them. I would wonder if anybody help me.

    a) If 5|3a+4b prove that 25|2a^2-3ab+2b^2
    b) If a^9|b^5 prove that a^7|b^4
    c) If a, b, c are natural numbers and a^b|b^a and b^c|c^b, prove that a^c|c^a
    a) is false. For example, if a=2 and b=1 then 3a+4b = 10 (multiple of 5), but 2a^2-3ab+2b^2 = 4 (not a multiple of 25).

    b) Suppose that p is one of the prime factors of a. Then p must also be a divisor of b. If p^s is the highest power of p that divides a, and p^t is the highest power of p that divides b, then the condition a^9|b^5 implies that 9s\leqslant5t, or \tfrac st\leqslant\tfrac59. Conversely, if that condition holds for each prime divisor of a, then a^9|b^5. The result then follows from the fact that \tfrac59<\tfrac47.

    c) Hint: notice that a^{bc} | b^{ac} | c^{ba}.
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  3. #3
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    Thanks for the help. The first problem was right but i had mistyped it

    a) If 5|3a+4b prove that 25|2a^2-3ab-2b^2
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    Quote Originally Posted by havaliza View Post
    Thanks for the help. The first problem was right but i had mistyped it

    a) If 5|3a+4b prove that 25|2a^2-3ab-2b^2
    In that case, notice that 2a^2-3ab-2b^2 = (a+2b)(2a-b), and show that each of those factors can be expressed as a multiple of 3a+4b plus some multiple of 5.
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