2
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$A_1$:

0H828JD1FGHODB82JOO

You're Pointless

$A_2$: 01111011

$A_3$: (Sorry if you are colorblind..)

Add this upside down.

$A_4$:

    83972748                                                                                                                                                                                                                                               & 8

$A_5$: (!(5%3)^4+7/2)*2


What is the result of the following calculation?

$~~~~~~~A_2 - A_1 + A_3 \times A_4 + A_5$

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  • 1
    $\begingroup$ In the last equation, I assume you replace each number with the corresponding answer? Also, should it come out to something meaningful? $\endgroup$ – Deusovi Mar 29 '16 at 12:37
  • $\begingroup$ @Deusovi I've added some info, but to answer your two questions: yes, replace each number with the number the individual item represents. And no, the answer is a number, but it doesn't have any other meaning that just being the result of the sum. $\endgroup$ – Kevin Cruijssen Mar 29 '16 at 12:56
  • $\begingroup$ I was thinking I am not colorblind but I saw that number really hardly. $\endgroup$ – Lafexlos Mar 29 '16 at 13:03
  • 1
    $\begingroup$ I've edited the question to make it clear which numbers were numbers and which were variables. Feel fre to roll back if you need to. $\endgroup$ – Deusovi Mar 29 '16 at 13:03
  • 1
    $\begingroup$ Why do you have to ask questions color blind people cannot answer? Tsk Tsk. $\endgroup$ – Marius Mar 29 '16 at 13:18
2
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  1. Not sure about this one yet.

  2. This is binary for 123. (ASCII produces {, which is not a number as far as I'm aware.)

  3. I'm colorblind, but the image description is "Add this one upside down". I think it's a 9, so the answer would be 6.

  4. There is a hidden & 8 after lots of spaces. Taking & as bitwise AND, we get 8.

  5. There is an equation hidden in the source; taking % as binary modulus, you get 39.

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  • $\begingroup$ Why are you converting 123 to ascii. I am new to this so want to make sure what am I missing. It dosa not say anywhere to convert $\endgroup$ – LearningPhase Mar 29 '16 at 12:41
  • $\begingroup$ Also same about the upside down number $\endgroup$ – LearningPhase Mar 29 '16 at 12:43
  • $\begingroup$ @LearningPhase: It doesn't say to, but numbers between 32 and 128 are ASCII. Binary suggested "computers" to me, so I tried ASCII. $\endgroup$ – Deusovi Mar 29 '16 at 12:45
  • $\begingroup$ @LearningPhase: The upside down number comes from the source of the question. Click "edit" on the question (but don't make any changes) and you'll see what I mean. $\endgroup$ – Deusovi Mar 29 '16 at 12:47
  • $\begingroup$ For the first I would go for 0 because, well, it's pointless in the formula. The rest is only a red herring (the rest is pointless) $\endgroup$ – Narmer Mar 29 '16 at 13:02
0
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  1. I would go for 0 because it's pointless in the formula. The rest is only a red herring (the rest is also pointless)

  2. This is binary for 123. (ASCII produces {, which is not a number as far as I'm aware.) Thanks to @Deusovi

  3. The image description is "Add this one upside down". Since it's a 9, the answer would be 6. Thanks to @Deusovi

  4. There is a hidden & 8 after lots of spaces. Taking & as bitwise AND, we get 8. Thanks to @Deusovi

  5. There is an equation hidden in the source; Also the third image points to a $2px*2px$ GIF. Pluggin them in the equation we get the total result of $[[((5\%3)^4+7/2)*2][2*2]][2*2] = 624$

So the final result is: $123 - 0 + 6*8 + 624 = 795$
I feel something's wrong...

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