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... 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
I'd say it's fifty-fifty.
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.
Sorry if this is stupid but how to deal with sums to odd numbers ? Won't you have a number left over after pairing all the others?
Add the last number onto the end. So the sum of all numbers between 1 and 101 is 50 pairs of 101 plus one extra 101 and the end. It'd end up being 5050 + 101 or 51x101 or 5151
Nope, because what you're doing is copying the entire sequence, reversing it, and pairing up each element left to right. There's no way to have any leftovers because the original sequence and the new reversed sequence have the same number of elements.
A perhaps less intuitive way of thinking of it is you start with a sequence of 1 up to N, which contains exactly N elements. The sequence from 1 to N and its reverse together contain 2N elements, which is by definition an even number, regardless of whether N is even or odd. Because it's even we can break it into pairs without leftovers.