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

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- Mar 21st 2007, 03:39 PMRimasSum of Positive Integers
Find the sum of all positive integers less than 1000 that are divisible by 3 but not by 2

- Mar 21st 2007, 04:02 PMThePerfectHacker
- Mar 27th 2007, 03:59 PMceasar_19134
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. - Mar 27th 2007, 11:09 PMDivideBy0
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 :cool:**

If you have any questions regarding the formula or anything else feel free to ask. - Mar 31st 2007, 04:34 AMSoroban
Hello, Rimas!

Quote:

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

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