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# Another Sum....

0
488
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What is the sum of the VALUES of all digits from 1 to 1,000,000? Just to make it clear. Example:

11+12+13+14+15+16+17+18+19+20=1+1+1+2+1+3+1+4+1+5+1+6+1+7+1+8+1+9+2+0=56...etc.

Thank you for any help.

Guest Apr 16, 2017
#1
+2

Here is my take on this one. Will do a twist on Gauss's method as follows:

0, 1, 2, 3..............999,996, 999,997,999,998, 999,999

Will add first and last=0 +999,999 =9 x 6 =54

Will add second to second last =1 + 999,998 =999,999 =9 x 6=54........and so on.

Since we have: 1,000,000/2 =500,000 pairs, each equaling 54, we therefore have:

500,000 x 54 =27,000,000 !!.

The only number left is the 1,000,000 itself, and since it has a value of 1, we therefore have:

27,000,000 + 1 =27,000,001 !!. And that is that, I THINK !!.

Guest Apr 16, 2017
#2
+310
+1

I didnt understand the guest's way, so ill solve it my way:

P(k)=the number of times the digit k appears between 0 and 999,999. P(0)*0+P(1)*1+P(2)*2+....+P(9)*9+1= the sum we need to find (i replaced 1,000,000 with 0, all we need to do is add 1 to the final solution.

all the numbers between 0 and 999,999 are also every possible combination of 6 digits (instead of 1 we write 0000001, instead of 2 we write 000002. Yes, a large number of zeroes is added, but that doesnt change the solution because x+n*0=x+0=x)

Here's a link to a similar question someone asked about the number of times a digit appears between 1 and 10n(n is a positive integer):

https://web2.0calc.com/questions/how-many-3s#r21

. i told him that between 0 and 10n-1 ,every digit (if we write it the way i write the numbers) appears n*10n-1 times, meaning every digit except for 0 appears n*10n-1 (but we dont care about the zeroes). what we want to find is the number of times every digit (except for 0) appears between 0 and 999,999=106-1. so, the answer is 6*106-1=6*105.

so, P(k)=6*105 for every k that is between 1 and 9.

P(0)*0+P(1)*1+P(2)*2+....+P(9)*9+1=P(1)*1+P(2)*2+....+P(9)*9+1=6*105(1+2+3+4....+9)+1=6*105*45+1=27,000,000+1=

27,000,001

Ehrlich  Apr 16, 2017
edited by Ehrlich  Apr 16, 2017
#3
+2

Here is another simple solution:

Will write all the numbers as 6 digits as follows:

000001, 000002, 000003...........999997, 999998, 999999.

Since we have 1,000,000 numbers, each of which is 6 digits, then:

Total number of INDIVIDUAL DIGITS =6 x 1,000,000 =6,000,000 digits.

Now, the average VALUE of each digit is:[0+1+2+3+4+5+6+7+8+9=45] / 10 =4.5

Therefore the total VALUE of the 6,000,000 digits will be:

6,000,000 x 4.5 =27,000,000. And will add 1 for the last of digit of 1,000,000 itself and will have:

27,000,000 + 1 =27,000,001 - Total Value of ALL digits from 1 to 1,000,000.

Guest Apr 16, 2017
#4
+92674
+1

CPhill  Apr 17, 2017