That seems complicated.
Average is (100+1)÷2= 50.5
50.5 average x 100 numbers = 5050.
You did essentially the exact same thing in a different order.
(100+1)÷2 × 100
(100+1)x100 ÷ 2
Clever though
The neat thing about math is that you can generally find several different paths to the same solution, and go with whichever is most intuitive to you.
Somewhat long story time.
Because I moved from a different country to the US they got confused what grade to put me in. And then they really goofed my class assignments. So I ended up literally only taking algebra 1 and geometry in high school. Unlike most that also got algebra 2 and possibly trig.
This set me up supery poorly for college. But I’m actually really good at math and I love math. So when I got to college, I failed miserably and dropped out after teachers were telling me this was simple stuff I should know from high school trig. How the college even let me get to that point was also insane…
Anyway years later I went back, they tested me and I needed remedial math classes. Finally this new college was doing it right. I immediately started failing my tests. The teacher actually cared to find out why. I was getting every answer correct, but I was only finishing about 50-60% of the questions. When they looked at my work they noticed I just want using the right formulas in the right places or the common shortcuts and such. At they focused on teaching me that stuff all of a sudden I was actually finishing my tests ahead of class and passing.
Math is weird.
Though sadly ime that is what teachers usually completely fail to convey. And then we wonder why so many people hate math.
I think you’d need to prove that the average is (100+1)/2 because that’s not an axiom.
… 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.
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Well, it does happen
IIRC, In this case the teacher tried to get smart ass Gauss shut up for a bit so he could teach the other students. It was only Gauss that had to solve the problem.
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…
I’m a spatial-visual person, so when presented with this problem as a teenager, I instead solved it spatially. If you stack squares like.
█.
██.
███.
…To the hundredth row, you get a shape that is a half filled square that is 100x100. Except the diagonal is fully filled in, so you need to add another 50.
So the answer was 0.5x100x100 + 0.5x100. Easy to visualize, easy to solve. 5050.
There’s a similar problem in sports – I was a teaching assistant for our rural school’s gym class so this one also popped up for me as a teenager. If you have 100 teams and each team needs to play each other team once… You fill in a similar grid, with the teams on both the x and y axis. The diagonal gets removed in this scenario because a team cannot play itself. So the answer is 0.5x100x100 - 0.5x100. 4950. Anyone who has ever tried to plan any sort of tournament can probably solve this intuitively, but 25 years ago I though I was the smartest gym class teaching assistant ever ;)
The algorithm gets a little weird if you’re summing the numbers to an odd number, though since there will be a left over number you have to deal with . By calculating 2S it works exactly the same in either case.
Ohhhhh
Is it required to wear a silly hat to be a genius mathematician? I’ve seen Euler and his hat. But I didn’t realize Gauss was in on it too.
Seeing this meme gives me flashbacks to the 10 Deutschmark bill (I think that was the one)
Just put it in the calculator.
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