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Math Help - 4 Inequality and number base problems

  1. #1
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    4 Inequality and number base problems

    The following math problems consists of two number base problems and two inequality problems that I have been unable to solve:

    1. A number X is converted into base 7 and becomes a four-digit number. Its leftmost digit is removed and written again as the rightmost digit. The number thus obtained in base 7 is twice X. Find the decimal representations of all such numbers X.

    2. Does there exist a positive integer which, when it is written in base 10 and its leftmost digit is crossed out, can be multiplied by 56 to get the original number?

    3. Two real numbers, a and b satisfy the equation ab = 1.
    (i) Prove that a6 + 4b6 >/= 4.
    (ii) Does the inequality a6 +4b6 > 4 hold for all a and b such that ab = 1?

    4. Numbers a, b and c are all greater than or equal to 0. Prove
    2(a3 + b3 + c3) >/= ab(a+b) + bc(b+c) + ca(c + a)

    Thanks a lot for answers to anyone of these questions
    Last edited by JeanMal; July 18th 2014 at 12:07 PM.
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  2. #2
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    Re: 4 Inequality and number base problems

    Hello, JeanMal!

    1. A number X is converted into base 7 and becomes a four-digit number.
    Its leftmost digit is removed and written again as the rightmost digit.
    The number thus obtained in base 7 is twice X.
    Find the decimal representations of all such numbers X.

    This is an alphametic in base-7:. \begin{array}{cccc} A&B&C&D \\ \times &&& 2 \\ \hline B&C&D&A \end{array}


    I found two solutions: . \begin{array}{cccc}1&2&5&4 \\ \times &&& 2 \\ \hline 2&5&4&1 \end{array}\;\;\text{ and }\;\;\begin{array}{cccc}2&5&4&1 \\ \times &&& 2 \\ \hline 5&4&1&2 \end{array}


    Therefore: . X \;=\;\begin{Bmatrix}1254_7 &=& 480 \\ 2541_7 &=& 960 \end{Bmatrix}
    Thanks from JeanMal
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  3. #3
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    Re: 4 Inequality and number base problems

    Hello, JeanMal!

    2. Does there exist a positive integer which, when it is written in base 10
    and its leftmost digit is crossed out, can be multiplied by 56 to get the original number?

    Let a be the leftmost digit.
    Let b be an n-digit integer.
    . . Then the number has the form: . N \:=\:10^na + b

    Crossing out the leftmost digit, we have b.
    And we want 56b to equal N.

    We have: . 56b \:=\:10^na + b \quad\Rightarrow\quad 55b \:=\:10^na

    . . . . . . . . 5\!\cdot\!11b \:=\:2\!\cdot\!5\!\cdot\!10^{n-1}a \quad\Rightarrow\quad11b \:=\: 2\!\cdot\!10^{n-1}a

    The equation holds if: . b = 2\!\cdot\!10^{n-1}\;\text{ and }\;a = 11

    But a = 11 is not a digit.

    The answer is: NO.

    Thanks from JeanMal
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  4. #4
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    Re: 4 Inequality and number base problems

    Quote Originally Posted by JeanMal View Post

    3. Two real numbers, a and b satisfy the equation ab = 1.
    (i) Prove that a6 + 4b6 >/= 4.
    (ii) Does the inequality a6 +4b6 > 4 hold for all a and b such that ab = 1?
    We have b=\displaystyle\frac{1}{a}.
    Replace b in the inequality: a^6+\frac{4}{a^6}\ge 4.
    Multiply both members by a^6: a^{12}+4\ge 4a^6\Leftrightarrow \left(a^6-2\right)^2\ge 0
    The equality holds for a=\displaystyle\sqrt[6]{2}.
    Then the second inequality isn't true for a=\sqrt[6]{2}, \ b=\displaystyle\frac{1}{\sqrt[6]{2}}
    Thanks from JeanMal
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  5. #5
    MHF Contributor red_dog's Avatar
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    Re: 4 Inequality and number base problems

    Quote Originally Posted by JeanMal View Post

    4. Numbers a, b and c are all greater than or equal to 0. Prove
    2(a3 + b3 + c3) >/= ab(a+b) + bc(b+c) + ca(c + a)
    Use the inequality a^2-ab+b^2\ge ab and multiply both members with (a+b).
    We get a^3+b^3\ge ab(a+b).
    In the same way b^3+c^3\ge bc(b+c) and c^3+a^3\ge ca(c+a).
    Then add the three inequalities.
    Thanks from JeanMal
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