... am I the only one who learned 1+100, 2+99... to make 101 times 50 pairs? Lmao feels like it's much easier. 101 × 50 = 5050
The math is the same, you just wrote it more "casually". For me it was 0+100, 1+99, 2+98 ... 49+51 -> 100 x 50 = 5000, then add the 50 that was missed from the middle for 5050. But yeah I remember coming up with that when I was really young.
This is my first time seeing this problem. Interesting that they taught it in school.
Had a statistics and probability class in hs instead of the standard precalc. I feel it's more applicable for students now than precalc anyways. It felt pretty cool to sit down in class and figure out the odds of winning on a lotto ticket and when the odds indicate you should buy a ticket.
Yeah pre-calc is pretty much remedial math nowadays. You don't even get 100 level math until you're at intermediate algebra!
Thinking of it in terms of statistics makes a lot of sense, I can see how this problem would help develop intuitions.
I always thought like that:
Hmmm: 1 + 2 + 3 + ... + 99 + 100
Kommutativgesetz be like: This equals:
100 +1 + 99 + 2 +98 + 3 . . . And this equals: 101+ 101+ 101+ . . .How often do I need to do this? I use up 2 numbers for each 101. I have 100 numbers total. So that's 50x101.
Now you can think about: What if it's 1000 instead of 100? But it#s easy from here...
Pythagoras ▄︻デ══━一 Gauss
Edit: Oops, that doesn't quite work without specifying why it obviously needs to be a 101 by 100 triangle, which I can't do on weed. Something about 1 side including 0-100 maybe?
