### Challenge Problem Showdown- April 22, 2013

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

=> z(x>6 and odd) = |x| – 3 – 2|x| + 8 + 2|x| – 12 – |x| + 7

=> z = 0 (everything gets cancelled out)

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 â€“ 3| â€“ 2|x â€“ 4| + 2|x â€“ 6| â€“ |x â€“ 7| = (3 â€“ x) â€“ 2(4 â€“ x) + 2(6 â€“ x) â€“ (7 â€“ x) =

= 3â€“8+12â€“7â€“x+2x-2x+x = 0

b) 3 <= x < 4

|x â€“ 3| â€“ 2|x â€“ 4| + 2|x â€“ 6| â€“ |x â€“ 7| = (x â€“ 3) â€“ 2(4 â€“ x) + 2(6 â€“ x) â€“ (7 â€“ x) =

= â€“3â€“8+12â€“7+x+2xâ€“2x+x = â€“6+2x

c) 4 <= x < 6

|x â€“ 3| â€“ 2|x â€“ 4| + 2|x â€“ 6| â€“ |x â€“ 7| = (x â€“ 3) â€“ 2(x â€“ 4) + 2(6 â€“ x) â€“ (7 â€“ x) =

= â€“3+8+12â€“7+xâ€“2xâ€“2x+x = 10â€“2x

d) 6 <= x < 7

|x â€“ 3| â€“ 2|x â€“ 4| + 2|x â€“ 6| â€“ |x â€“ 7| = (x â€“ 3) â€“ 2(x â€“ 4) + 2(x â€“ 6) â€“ (7 â€“ x) =

= â€“3+8â€“12â€“7+xâ€“2x+2x+x = â€“14+2x

e) 7 6 the value cannot be determined because for x from 6 to 7 we get different values of â€“14+2x for different x, e.g. if x=6.5 an expression equals to â€“1, but for x=7 it is 0.