Thread: Sum of Positive Integers

Find the sum of all positive integers less than 1000 that are divisible by 3 but not by 2

Originally Posted by Rimas

Find the sum of all positive integers less than 1000 that are divisible by 3 but not by 2

3,6,9,12,15,....,999
How many? Answer (333)

Now list all those divisible by 2:
6,12,18,....,996
How many? Answer (83)

Subtract them to get answer.

Rimas and I are friends, and you seem to have read the problem wrong like I did intially. The problem asks for the sum of the integers, not how many integers.

I already determined that the sum of all integers divisible by 3 and 2 is 82,170 if that helps any.

To solve this problem, you have to know this formula: n/2 * (a + L), where a is the first number in the series, L is the last number in the series, and n is the number of numbers in the series. It gives the sum of the series.

So... you have to sum every multiple of 3 first. As there are 333 of them, the first is 3, and the last is 999, sub that in to find the sum:

333/2 * (3 + 999) = 166833

Next, subtract all the number divisible by both 2 and 3 (that is, numbers divisible by 2*3 = numbers divisible by 6). As there are 166 of them, the first is 6 and the last is 996, sub that in to find the sum:

166/2 * (6 + 996) = 83166

Now, simply subtract 83166 from 166833:

166833 - 83166 = 83667

If you have any questions regarding the formula or anything else feel free to ask.

Hello, Rimas!

Find the sum of all positive integers less than 1000 that are divisible by 3 but not by 2

The numbers divisible by 3 are: .3, 6, 9, 12, 15, 18, ... , 999

We see that every other number is divisible by 2, so we omit those.

And we have: .3, 9, 15, 21, ..., 999, .an arithmetic sequence
. . with first term a = 3, common difference d = 6, and n = 167 terms.

The sum of the series is: .½(167) [2(3) + 166(6)] .= .83,667