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OG 2016 #87 PS

This topic has 2 expert replies and 1 member reply
Azizakaria Senior | Next Rank: 100 Posts
Joined
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Posted:
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OG 2016 #87 PS

Post Fri Oct 09, 2015 9:41 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    87. The sum of all the integers k such that -26 < k < 24 is
    (A) 0
    (B) -2
    (C) -25
    (D) -49
    (E) -51

    i don't understand why the answer is D

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    Post Fri Oct 09, 2015 9:47 am
    Hi Azizakaria,

    We're limited to the INTEGERS that fall in the range -26 < K < 24....so that would be....

    -25, -24, -23.....0.....22, 23

    Each of the positive integers form 1 to 23 (inclusive) would be 'cancelled out' by the negative integers from -1 to -23 (inclusive). The sum of ALL of those terms would be 0. But then you have to consider the other remaining integers: -24 and -25. Adding those two numbers gets us -49.

    Final Answer: D

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    Rich

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    Thanked by: Reyad2016
    Post Fri Oct 09, 2015 9:51 am
    Azizakaria wrote:
    87. The sum of all the integers k such that -26 < k < 24 is
    (A) 0
    (B) -2
    (C) -25
    (D) -49
    (E) -51

    i don't understand why the answer is D
    (-25) + (-24) + (-23) + (-22) + (-21) + . . . . .+ 21 + 22 + 23

    Notice that the blue values all cancel out, because for each positive value, there's a negative value with the same magnitude (e.g., -23 + 23 = 0, and -22 + 22 = 0, etc)

    So, we're left with (-25) + (-24), which equals -49

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    Azizakaria Senior | Next Rank: 100 Posts
    Joined
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    Posted:
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    3 times
    Post Sat Oct 10, 2015 2:44 am
    now I understand, thanks for your help, the explanation in the OG is so unclear.

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