Question 236340: What is the smallest integer that could be part of a set of two or more positive consecutive integers whose sum is 100?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! I believe the number is 9
9 sum 9
10 sum 19
11 sum 30
12 sum 42
13 sum 55
14 sum 69
15 sum 84
16 sum 100
I summed the numbers from 1 up to when i got past 100.
I then looked at each number over 100 to see if there were sums in the lower range of numbers that I could subtract to get 100.
When I got to the sum of the number at 16 was 136, I saw that the sum of the numbers at 8 totaled 36 so I could subtract the first 8 numbers from the set and my total from 9 to 16 would equal 100.
It worked although I don't have a formula to apply to it.
The formula for the sum of an arithmetic progression can be applied.
It is:
S = n*(x+y)/2 where
x = first number in the sequence
y = last number in the sequence.
For this progression, we have:
n = 8
x = 9
y = 16
the formula becomes:
S = 8*(9+16)/2 which becomes:
S = 8*25/2 = 100
Our solution became:
S = 16*(1+16)/2 - 8*(1+8)/2 which became:
S = 8*17 - 4*9 = 136 - 36 = 100
I suspect there's a generic formula in there to help solve this problem a little more rigorously using algebra, but I don't have that answer for you.
The answer, however, is correct as best as I can determine, and it is:
The smallest number in a consecutive series that adds up to 100 is 9.
The number of numbers involved is 8.
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