Question

Could someone please help me solve this problem?

Answer 1

solution In the ones place, we run through the digits 0−9 ten times, once for each decade. One run of the digits 0−9 adds up to 45, and so the ones place contribues 10⋅45=450 to the sum.

In the tens place, each of the digits occurs ten times, once for each of the numbers in its decade (for example, we get ten 2s: one for each of the numbers between 20 and 29). So the tens place contribues another 450 to the sum.

The hundreds place contributes only 1 more, so the total sum is 901.

Answer 2

it consumes more time @Directrix

Answer 3

@sunil186 The chart does consume more time but is instructive for those who do not know how to get started on the problem.

And, I read base 10 as base 8, so I messed up. I now agree with you on the 901.

Answer 4

i appreciate @ directrix

Answer 5

@sunil186 I am thinking now that you may have answered the attached question. See what you think.

Answer 6

yeah i did answered attached question bt it was for base 10 @ directrix

Answer 7

The numbers are written out in base 8 so how could there be a digit 9 >> One run of the digits 0−9? @sunil186

Answer 8

no the numbers are written out in base 10 see the question properly ..@ directrix

Answer 9

I think one run (sum of a row) of the digits in base 8 would be 28. Just thinking.

Answer 10

Am I looking at the wrong problem. It is stated that the numbers are written out in base 8.