There are 100 integers from 0 to 99, inclusive.
Considering 0 to have no digits, the sum of all integers with none, 1 or 2
digits is 

Put a 0 on the end of each, and you've multiplied each by 10, and now the sum
of those 100 is 10 times as much, or 49500, and each of those 100 now ends
with 0.
 
Add 3 to each of those 100, and you've added 300 to the 49500 sum, so there are
49500+300=40800 that end with 3.

Add 4 to each of those 100, and you've added 400 to the 49500 sum, so there are
49500+400=40900 that end with 4.

Add 5 to each of those 100, and you've added 500 to the 49500 sum, so there are
49500+500=50000 that end with 5.

Add 8 to each of those 100, and you've added 800 to the 49500 sum, so there are
49500+800=50300 that end with 8.

Total = 49800+49900+50000+50300 = 200000

Edwin

I'm going to do more work than the problem requires, to make sure I'm doing

it correctly, and hopefully this method helps you gain some intuition.



Sum of all positive integers less than 1000 ending in 0:

Sum(0+10n; n=0,...,99) = 0*sum(0, n=0,...,99) + 10*sum(0...99) = 10*(99*100/2) 

  = 0 + 49500 = 49500



Sum of all positive integers less than 1000 ending in 1:

Sum(1+10n; n=0,...,99) = 1*sum(1, n=0,...,99) + sum(10n, n=0,...,99)

 = 100 + 49500 = 49600



Repeating the above two calculations and noting we just add 100 for each

higher digit:



Sum of all positive integers less than 1000 ending in 0:   49500

Sum of all positive integers less than 1000 ending in 1:   49600

Sum of all positive integers less than 1000 ending in 2:   49700 

Sum of all positive integers less than 1000 ending in 3:   49800 (#)

Sum of all positive integers less than 1000 ending in 4:   49900 (#)

Sum of all positive integers less than 1000 ending in 5:   50000 (#)

Sum of all positive integers less than 1000 ending in 6:   50100

Sum of all positive integers less than 1000 ending in 7:   50200

Sum of all positive integers less than 1000 ending in 8:   50300 (#)

Sum of all positive integers less than 1000 ending in 9:   50400

                                                  total:  499500



Sanity check the above total, which should be the same as ALL numbers 0...999:

Sum(n; n=0,...,999) = 999*1000/2 = 499500   



Sum the rows with (#), you should get 200000

 

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!
 








...




The sum of all of these gives us an arithmetic sequence of 100 terms with first term 20 and last term 3980.


Using the informal formula for the sum of the terms of an arithmetic sequence


Sum = (# of terms) * (average of first and last)


gives us





ANSWER: 200,000


 

Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!
 

This is a slight variation of what tutor greenestamps did.





3+4+5+8 = 20

Add 10 to each item on the left side, so you're adding 40 to both sides

13+14+15+18 = 60

Repeat the previous step

23+24+25+28 = 100

Repeat again

33+34+35+38 = 140

And so on.





The final four terms would add to 993+994+995+998 = 3980





The right hand sides of those equations form this arithmetic sequence 20, 60, 100, 140, ..., 3980





The first terms of each left hand side are 3,13,23,33,...,993. 

Erase the units digit to arrive at 0,1,2,3,...,99

This shows we have 99-0+1 = 100 equations and therefore we have 100 terms in the sequence 20, 60, 100, 140, ..., 3980





Sn = sum of first n terms of arithmetic sequence

Sn = (n/2)*(firstTerm + nthTerm)

S100 = (100/2)*(20 + 3980)

S100 = 200,000 is the final answer.

The comma separator is there to make the single number more readable.

Erase the comma if needed.





Another way to arrive at this answer is to say,

Sn = (n/2)*(2*a + d*(n-1))

S100 = (100/2)*(2*20 + 40*(100-1))

S100 = 200,000

 

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