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This may be a rather silly question, but I found myself thinking about the famous anecdote of Gauss summing from 1 to 100 in school, where he sums $$1+100=101, 2+99=101,...$$, and so on. He concludes that there are 50 pairs of numbers between 1 and 100 that sum 101, so the answer is that the first 100 natural numbers sum to $$50*101=5050$$ So I tried arriving at the same result in a similar way by going 0+100=100, 1+99=100,... and so on, but this gives me 51 pairs (since we must consider 0 now) summing 100, making the answer $$51*100=5100.$$This is obviously wrong but i can't find the error.

J. W. Tanner
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Lorentz
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    Did you count $50$ twice, because $50+50=100$? – J. W. Tanner Jul 20 '23 at 21:36
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    You get $50$ pairs $0+100,\dots,49+51,$ and a lonely number $50$ – Anne Bauval Jul 20 '23 at 21:36
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    In general, it is very helpful when doing this "pairwise" summing up to watch out for an odd number of terms; special cases must be dealt with specifically(like the single $50$) – dgeyfman Jul 20 '23 at 21:47
  • Heh, when I was in fourth grade, I was given this problem. I was told there was a trick, and found this $0+100,\dots$ approach. But I figured out to leave the $50$ case separate – Thomas Andrews Jul 20 '23 at 22:57
  • You can avoid this by writing the sum twice , first from $1$ to $100$ , then in reverse. Then, we have $100$ pairs summing up to $101$. Although this is considered as an anecdote , I see no reason why Gauss could not have had this idea as a youngster. Some sources claim he was the first , hard to verify. – Peter Jul 21 '23 at 00:03

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I suspect the error is that you counted $50$ twice, because $50+50=100$,

and that is why you came up with an answer that is $50$ more than the correct one.

J. W. Tanner
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