Challenge Problem Showdown- April 22, 2013

Lauren Golin —  April 22, 2013 — 2 Comments

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Here is this week’s problem:

If x is positive, what is the value of |x “ 3| “ 2|x “ 4| + 2|x “ 6| “ |x “ 7| ?

(1) x is an odd integer.

(2) x > 6

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Lauren Golin

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2 responses to Challenge Problem Showdown- April 22, 2013

  1. Start from the definition of |Z|.
    |Z| = Z if Z >= 0 and
    |Z| = -Z if Z < 0
    Now consider the following cases:
    a) x < 3
    |x – 3| – 2|x – 4| + 2|x – 6| – |x – 7| = (3 – x) – 2(4 – x) + 2(6 – x) – (7 – x) =
    = 3–8+12–7–x+2x-2x+x = 0
    b) 3 <= x < 4
    |x – 3| – 2|x – 4| + 2|x – 6| – |x – 7| = (x – 3) – 2(4 – x) + 2(6 – x) – (7 – x) =
    = –3–8+12–7+x+2x–2x+x = –6+2x
    c) 4 <= x < 6
    |x – 3| – 2|x – 4| + 2|x – 6| – |x – 7| = (x – 3) – 2(x – 4) + 2(6 – x) – (7 – x) =
    = –3+8+12–7+x–2x–2x+x = 10–2x
    d) 6 <= x < 7
    |x – 3| – 2|x – 4| + 2|x – 6| – |x – 7| = (x – 3) – 2(x – 4) + 2(x – 6) – (7 – x) =
    = –3+8–12–7+x–2x+2x+x = –14+2x
    e) 7 6 the value cannot be determined because for x from 6 to 7 we get different values of –14+2x for different x, e.g. if x=6.5 an expression equals to –1, but for x=7 it is 0.

  2. If x is positive, what is the value of |x – 3| – 2|x – 4| + 2|x – 6| – |x – 7| ?

    (1) x is an odd integer.

    (2) x > 6

    Ans:-
    Since, x is odd and greate than 6 => all mod values has to be positive and it can be expanded as below:

    let Z(x)= |x – 3| – 2|x – 4| + 2|x – 6| – |x – 7|
    => z(x>6 and odd) = |x| – 3 – 2|x| + 8 + 2|x| – 12 – |x| + 7
    => z = 0 (everything gets cancelled out)

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