One to One Million - Numberphile

Puzzles, classroom stories and the great Carl Gauss - oh, and adding every digit in the numbers between one and one million!
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NUMBERPHILE

Videos by Brady Haran

[Nessa-]

"I'm gonna skip a few"
skips 999 991 numbers

[flinkstiff]

I wrote all the numbers 0 - 1 000 000 on paper and added it all up. It took half a year. Would have gone faster but i had to sleep as well. I would say your method is superior to mine. Good thing is i had a lot of paper to burn for warmth this winter. My house were cold though because of me spending all my money on buying paper to solve this equation. I guess it was a paradoxical problem and solution.

[Niosus]

Another simple way is by looking at how often each number happens in each position. From to It is obvious that all possible combinations of numbers occur, so all you need is the average value of each position. 4.5 is the average off the numbers 0 through 9. You have 6 positions, so 4.5*6 = 27 is the average value of your numbers. For a million numbers that gives 27 million, with the last 1 for the 1 million.

This trick works on any arbitrary range with some care. For instance from 0 to For the last 5 digits, the old trick still works. For the first digit you need to average the range from 0-5 which is 2.5. That makes 5*4.5 for the last 5 positions (=22.5) and 2.5 for the first position to make an average of 25. 25 * = 15 million, plus the final 6 makes 15 000 006.

I believe this method is quite a bit more flexible and easier to calculate since you're working with some very small and easy to comprehend numbers right up until the last step.

[shelverman]

I'm a teacher, and I've been looking all over YouTube for a good telling of the story of Gauss. Thank you!

[jyak]

hmmm well 1 10 100 1, 000 10, 000 100, 000 and 1, 000, 000 all equal 1. so the answer is at least 7.

[Manabender]

I got the right answer quickly, but with a different method.

(More text to ensure a "Read more" button)

(A little more)

I considered six "digit slots", for which list all possibilities. Over ten numbers, the rightmost slot cycles through each digit 0~9, the sum of which is 45. This process repeats 100, 000 times.

Since each digit is independent, and all possibilities are used, each digit will appear 100, 000 times in each slot. The sum of each ten is 45, there are 6 slots, and there are 100, 000 groups of ten. 6*45*100, 000=27, 000, 000.

And of course, chuck on the +1 from 1, 000, 000, for 27, 000, 001

[Cloiss_]

The answer is 27, 000, 001 because the digits from 1 to 9 add up to 45, and each occur times in each of the 6 digits spots, so and then add 1 for 1, 000, 000.

[Kwailah]

This is the first time in a long time that I can remember someone explaining math and at the end I thought "oh that's so cool" 😊

[tanukigalpa]

I think I was taught this trick a long time ago in one of my math classes, but I still was so pleased when I could remember "pairs, pairs!". It really is a testament to our brains and the power of our memories that a problem explained for 10 minutes sometime in the last 10 years could be recalled within minutes today. I didn't work it out myself, but the explanation stuck with me!

[Misterspork]

I came about a different method, though not nearly as fast but it was still fun working out.

I considered that the sum from 1 to 9 of the digits was 45. In continuing this to 1-99 that sum was repeated 10 times in the ones digit and 10 times in the tens digit. The sum is also repeated 100 times in each the ones, tens, and hundreds digit in the sum from 1 to 999, and so on. This means the sum of 1-9 is repeated times in the sum in each digit. finally the +1.

[sphakamisozondi]

01:40 Shoutout to Dr James for roasting Gauss's maths teacher. 🤣🤣

[williamkristian]

Why do numberphile's thumbnails look like the funniest memes . Like a huge equation and it just zooms in on some old painting, with distortion

[christosvoskresye]

I thought you would do this a different way.  Adding the digits 0+1+2+3+4+5+6+7+8+9=45, and from 0 to 999, 999 has each of these digits appearing 100, 000 times in each column, so those digits add to 45 X 100, 000 X 6 = 27, 000, 000.  Then, once again, add the 1 from 1, 000, 000.

[lemonke]

I did it a different way to get the answer, i used a very complicated algorithm that some refer to as google

[Parasmunt]

A great example of how in maths problems can become incredibly easier when you just write them differently.

[VENMXVI]

I love it how enthusiastic he is in his videos! Great!

[AnnaGlin]

The film (based on the book) "Measuring the World" beautifully illustrates this story right at the beginning :)
It's a german film about the lives of Carl Friedrich Gauß and Alexander von Humboldt. I can recommend it (don't know about an english dub though)

[Jimpozcan]

To make it easier go from 0 to  If you include leading zeros there'll be no effect on the sum but each number will have six digits.  The average value of a digit is 4.5.  A value of 4.5 per digit times six digits per number times a million numbers from to is 27 million.  Add 1 for and you get

I think that this method is a little easier but it's not really that different.

[TylerWare]

Another fun way to do this problem is to consider the number of times the sum 1-9 (value of 45) can occur. From 0-9 it occurs once. From 0-99 it occurs 20 times. From 0-999, 999 it occurs 600, 000 times. From that you can develop a simple formula that gives 45*6*10^5 = 27, 000, 000 => +1 => 27, 000, 001. More generally, for any number that can be expressed as 10^n (where n is a natural number): 45*n*10^(n-1) + 1

[benl]

Consider the last digit of every number from 1 to 1 million. 1/10 of those digits will be 0, 1/10 will be 1, 1/10 will be 2, all the way up to 9. Thus, their average will be (0+1+2...+8+9)/10=4.5. The same is true of the second last digits, the third last digits, all the way up to the sixth last digit. For the seventh digit, it will only be 1 in a single case Thus, the sum of all the digits is