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• Day 5: Mental Math Warm-Up
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• Day 5: Easy as 1-2-3
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  @rz923 That's an interesting thought! If you'd encountered this problem on a math competition and, say, didn't have enough time to think of the way Prof. Loh demonstrated in the video, I'd say this would work!

  It's not completely rigorous, though. I definitely agree with you that you can find the first 3 digits by doing 369 + 36 + 3 + (whatever # is carried into the 3rd column), but how do you know for sure that that carried # is 2? For example, if you actually were to write out the whole sum of 369369369's, you'd see that the carried digit, from right to left, starts as 1, but goes all the way up to 5, and then down to the 2 you mentioned. It is a very reasonable guess that the 3rd column carried digit is 2-- it definitely cannot be one because the 4th column has at least 3+6+9+3 = 21. And the only way that the digit carried into the 3rd column could be 3 is if the digit carried into the 4th column is 9. This isn't really likely.

  Basically, the way you proposed is definitely valid, but if you wanted to be 100% sure, you could also use the other method.

  Hope that made sense! 🙂

• Day 5: A Man and a Trip
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  @professionalbronco That's close! The harmonic mean is actually equal to 2/(1/x + 1/y), or 1/(1/(2x) + 1/(2y)).

  To make it a bit clearer why the average speed is the harmonic mean, we can derive it from the beginning...

  So again, we go half the distance at x km/hr and half at y km/hr. Let's say that the total distance is k km. Then, we go (k/2) km at the speed x km/hr and (k/2) km at the speed y km/hr.

  And we want to find the average speed-- well, that's just the (total distance)/(total amount of time). We already know (total distance) = k, so let's find (total amount of time): it's equal to (total amount of time going the first k/2 km) + (total amount of time going the second k/2 km).

  Well, for the first k/2 km, we see that it takes (k/2) km * hr/x km = k/(2x) hr. And the second k/2 km takes k/(2y) hr. Meaning, we have a total time being k/(2x) + k/(2y).

  Finally, the average speed is k/(k/(2x) + k/(2y)) = 1/(1/(2x) + 1/(2y)) = 2/(1/x + 1/y), or the harmonic mean of x and y.

  So that's another way to figure it out, besides the faster but maybe a little less intuitive way Prof. Loh brings up in the video. The good thing is, once you understand how it's derived, then you can apply it without deriving every single time 😄

  Hope this helped!
  (Also, as a general sidenote: you can take the harmonic mean of more than 2 numbers-- in which case, the harmonic mean equals (the # of #'s)/(the sum of all the reciprocals of the numbers), which does come out to be 2/(1/x + 1/y) for 2 terms)