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Question 1150957: Find the sum of all positive three-digit numbers divisible by 5
Found 3 solutions by greenestamps, rothauserc, Edwin McCravy:
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
These numbers form an arithmetic sequence: 100, 105, 110, ..., 990, 995.
The sum of the terms of an arithmetic sequence is the number of terms, multiplied by the average of the terms.
(1) number of terms....
There are 900 3-digit numbers (100 to 999); one out of every 5 is divisible by 5. So the number of terms in this sequence is 900/5 = 180.
(2) average of the terms....
The average of the terms of an arithmetic sequence is the average of the first and last. So for this sequence the average is (100+995)/2.
(3) the sum....
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! The smallest three-digit number divisible by 5 is 100 and the largest is 995
:
Therefore, this is an arithmetic sequence - the formula for the sum of an arithmetic sequence is
:
Sum of first n terms = (n/2) * (a + L), where a is the first term and L is the last term
:
the formula for the nth term is
:
a(n) = a +d * (n-1), where a is the first term and n is the common difference
:
995 = 100 +5 * (n-1)
:
995 = 100 + 5n -5
:
5n = 900
:
n = 180
:
sum of first n terms = (180/2) * (100 + 995) = 90 * 1095 = 98550
:
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the sum of all positive three-digit numbers divisible by 5 = 98,550
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:
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website! Another way:
100 + 105 + 110 + … + 985 + 990 + 995
First we find out how many terms there are. To
do that, we make a new sequence by subtracting
100 from each term and writing what you get
under each term:
100 + 105 + 110 + … + 985 + 990 + 995
0 + 5 + 10 + … + 885 + 890 + 895
Divide every term through by 5 and write what
you get under each term:
100 + 105 + 110 + … + 985 + 990 + 995
0 + 5 + 10 + … + 885 + 890 + 895
0 + 1 + 2 + … + 177 + 178 + 179
So we see there are 180 terms, counting the 0
at the first.
Now we write the original series:
100 + 105 + 110 + … + 985 + 990 + 995
Underneath it we write the sum of the same series in reverse
100 + 105 + 110 + … + 985 + 990 + 995
995 + 990 + 985 + … + 110 + 105 + 100
Now add all the terms together and place the sums underneath:
100 + 105 + 110 + … + 985 + 990 + 995
995 + 990 + 985 + … + 110 + 105 + 100
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1095 +1095 +1095 + … +1095 +1095 +1095
Since there are 180 terms, that last sum is 1095×180 = 197100
But that is twice the desired term since we added it twice,
once forward and once backward. So we divide 197100 by 2 and
we get what the other tutors got:
197100÷2 = 98550 <-- answer
Edwin
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