Question 1209367: Find the sum of all positive integers less than 1000 ending in 3 or 4 or 5 or 8.
Found 4 solutions by mccravyedwin, math_helper, greenestamps, math_tutor2020:
Answer by mccravyedwin(407)   (Show Source):
You can put this solution on YOUR website!
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
Answer by math_helper(2461)  (Show Source):
You can put this solution on YOUR website!
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):
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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