+0

# Sum of.....

0
752
4

What is the sum of all digits from 1 to 1,000. Just to make it very clear, the question is about the sum of each individual number. Example: 11 =1 + 1 =2, 12= 1 + 2 =3..........97=9 +7 =16.....991 =9+9+1=19......and so on from 1 to 1,000. Thanks for help.

Mar 14, 2018

#1
+1

dont have the time to do them all, but every single one of your insane amount of numbers has a digit sum between 1 and 27.

8)

Mar 14, 2018
#2
+1

Well, you could add them up like this:

000 + 001 + 002 + 003..................996 + 997 + 998 + 999

Add the first and last: 000 + 999 = 27

Add second to second last =001 + 998 =27

Add third to the third from last =002 + 997 =27.......and so on for a total of: 1,000/2 =500

500 x 27 + 1[the last digit of 1,000 itself] =13,501 - which is the sum of all digits from 1-1,000.

Mar 14, 2018
#3
0

you can also do this way as "OfficialBubbleTanks" suggested:

Take the average of 0 + 27 =27/2 =13.50. And since you have 1,000 numbers, then:

13.5 x 1,000 + 1 =13,501, which is the same as above answer.

Mar 14, 2018
#4
+1

What is the sum of all digits from 1 to 1,000. Just to make it very clear, the question is about the sum of each individual number.

Example: 11 =1 + 1 =2, 12= 1 + 2 =3..........97=9 +7 =16.....991 =9+9+1=19......and so on from 1 to 1,000.

$$\text{Let  b =  numeral system } \\ \text{Let  s =  sum of all digits from 1 to b^n }$$

$$Formula:\\ \begin{array}{|rcll|} \hline s &=& 1 + n\cdot b^n \cdot \left( \dfrac{b-1}{2} \right) \\ \hline \end{array}$$

$$Example:\\ \begin{array}{|rcll|} \hline b &=& 10 \\ n &=& 3 \\ \text{from 1 to 1000 } : \\ s &=& 1 + 3\cdot 10^3 \cdot \left( \dfrac{10-1}{2} \right) \\ &=& 1 + 3000 \cdot \left( \dfrac{9}{2} \right) \\ &=& 1 + 13500 \\ &=& 13501 \\ \hline \end{array}$$ Mar 16, 2018
edited by heureka  Mar 16, 2018