math problem

If a book was numbered from 1 to 100, what would the sum of the digits be?

EX: 100 would be 1+0+0=1 and 99 would be 9+9=18

lol

it is 1,2,3,4,...96,97,98,99,100

so its every # from 1-100

Hmm 541 innit?

We can take the average of all values (5), multiply by the number of values (9, excluding zero which has no effect) and multiply by the number of times each value appears between 1 and 99. Then the 1 from 100 is added since it isn't already figured in. 5 · 9 · 12 + 1 = 541

901 I would guess.

Adding all the last digits would be like (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) ten times, then + 10 (there are 10 numbers that have a ten's digit of 10) + 20 (there are 10 numbers that have a ten's digit of 2) + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 1 (for 100).  That's 10(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) + 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 1, so that's 901.

wow too many numbers ...

way too many numbers ...

you just need to put a ! right?, no that's multiplication.... I don't think there's something for that....

There's your head...