11
$\begingroup$

Please convince them otherwise and save me from going into the dustbin!

I calculate numbers as below because I was programmed this way. But people don't understand me, and I'm on the verge of being thrown away. Please tell them why I calculate the way I do.

{The math symbols used work just like you normally expect them to.}


  • 750 + 300 = 11
  • 630 * 420 = 18
  • 990 / 1710 = 1
  • 870 - 270 = (-4)
  • 60 * 600 = 16
  • 510 + 1230 = 10
  • 2160 - 780 = 10
  • 660 / 1110 = 10
  • 1140 * 150 = 10

Some of the least complicated calculations:

  • 30 + 60 = 3
  • 180 + 540 = 12

The numbers on the left side of the equations need to be converted into other numbers somehow.

My master programmed me while watching professional snowboarders executing various jumps.

$\endgroup$

1 Answer 1

12
$\begingroup$

To solve this, you need to first

treat the numbers on the left side as degrees,

and then

apply them to a clock face where 0 degrees is 12, 30 degrees is 1, 60 degrees is 2, 90 degrees is 3, etc.

Keep in mind that

you're basically in mod 360 because the clock has 360 degrees.

So to show how a few work:

30 (1 o'clock) + 60 (2 o'clock) = 3
180 (6 o'clock) + 540 (540mod360=180→6 o'clock) = 12
750 (750mod360=30→1 o'clock) + 300 (=10 o'clock) = 11
630 (630mod360=270→9 o'clock) * 420 (420mod360=60→2 o'clock) = 18
990 (990mod360=270→9 o'clock) / 1710 (1710mod360=270→9 o'clock) = 1

$\endgroup$
1
  • 2
    $\begingroup$ To Dan Russell: Perfect rationale! Leaves nothing to be desired! $\endgroup$
    – Segwayinto
    Commented Jun 19, 2016 at 9:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.