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Math Help - Number line problem

  1. #1
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    Number line problem

    On a number line, tick marks are spaced so that the distance from a to b is twice the distance from b to c and the distance from b to c is twice the distance from c to d. If a, b, c, and d are positive integers, what is the smallest possible value of d?

    How can we show that it is 8?
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  2. #2
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    Re: Number line problem

    Quote Originally Posted by mjoshua View Post
    On a number line, tick marks are spaced so that the distance from a to b is twice the distance from b to c and the distance from b to c is twice the distance from c to d. If a, b, c, and d are positive integers, what is the smallest possible value of d?

    How can we show that it is 8?
    1. The smallest possible tick is at 1. Thus a = 1.

    2. Let x denote the distance between a and b: b = a + x

    3. Then you get:

    \displaystyle{d = a + x + \frac12 x + \frac12 \left(\frac12 x \right) = 1 + \frac74 x = \frac{7x+4}4}

    4. Obviously the quotient is a positive integer if x = 4, therefore d = 8.
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  3. #3
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    Thought twice

    Quote Originally Posted by mjoshua View Post
    On a number line, tick marks are spaced so that the distance from a to b is twice the distance from b to c and the distance from b to c is twice the distance from c to d. If a, b, c, and d are positive integers, what is the smallest possible value of d?

    How can we show that it is 8?
    Could it be that you didn't mention that a < b < c < d ?

    Since the distance is a scalar (and not a vector) the ticks could be arranged:
    Code:
    0--1--2--3--4--5\\
    ---a--d--c-----b
    In this case the smallest value of d = 2.

    Maybe there other arrangement possible (?)
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  4. #4
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    Re: Number line problem

    Hello, mjoshua!

    Did you make a sketch?


    On a number line, tick marks are spaced so that:
    . . the distance from a to b is twice the distance from b to c
    . . and the distance from b to c is twice the distance from c to d.
    If a, b, c, and d are positive integers, what is the smallest possible value of d?

    How can we show that it is 8?

    The number line looks like this:

    . . \begin{array}{cccccccc} & 4x && 2x && x \\ \bullet & ----- & \bullet & --- & \bullet & -- & \bullet \\ a && b && c && d \end{array}

    Since we want the smallest value of d, let x \,=\,\overline{cd} \,=\,1
    . . and let a = 1, the smallest positive integer.


    Then we have:

    . . \begin{array}{cccccccc} & 4 && 2 && 1 \\ \bullet & ----- & \bullet & --- & \bullet & -- & \bullet \\ 1 && 5 && 7 && 8 \end{array}


    Therefore: . d\,=\,8



    Earboth,
    I love your "second thought"!
    And I'm impressed that you found a solution . . .
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  5. #5
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    Re: Number line problem

    Quote Originally Posted by Soroban View Post
    [size=3]...

    Earboth,
    I love your "second thought"!
    And I'm impressed that you found a solution . . .
    Soroban,

    Thank you very much.

    Even though it isn't necessary - I believe - I will only state for completeness that d = 4 is of course a valid result too.
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