a - What is the total number of individual or separate digits from 1 to (2^18 + 3^10) =321,193? Example. 1 to 9 =9 digits. 10 to 20 =2 x 10 =20 digits. And so on.
b - How is this total distributed among the ten digits 0 to 9. Or, how many 0s, 1s, 2s, 3s, 4s, 5s, 6s, 7s, 8s, 9s does the total in (a) above contain?
c - What is the total sum of all the digits calculated in (a) above?
Any help would be appreciated. Thank you.
a -
9 x 1 = 9
90 x 2 = 180
900 x 3 =2,700
9,000 x 4 =36,000
90,000 x 5 =450,000
321,193 - 100,000 + 1 =221,194
221,194 x 6 =1,327,164
So: 9 + 180 + 2,700 + 36,000 +450,000 + 1,327,164 =1,816,053 - digits in total.
b -
This above number of 1,816,053 is distributed among the ten digits of 0 to 9 as follows:
(0s =158,329, 1s =268,628, 2s =259,534, 3s =179,534, 4s =158,339, 5s =158,339, 6s =158,339, 7s =158,339, 8s =158,339, 9s =158,333 = 1,816,053) - which balances with the total in "a" above.
P.S. These were calculated by a simple computer code written for this purpose.
c -
To sum up all the above digits, proceed as follows:
We simply take the breakdown in "b" above, multiply them by their respective values and add them all up:
158,329*0 + 268,628 *1 + 259,534*2 + 179,534*3 + 158,339*4 + 158,339*5 + 158,339*6 + 158,339*7 + 158,339*8 + 158,333*9 =7,501,465 - sum of all the above 1,816,053 digits in "a".
