Multiples of $3$ or $5$

If we list all the natural numbers below $10$ that are multiples of $3$ or $5$, we get $3, 5, 6$ and $9$. The sum of these multiples is $23$.
Find the sum of all the multiples of $3$ or $5$ below $1000$.

To tackle this problem, we should first find the individual sums of the multiples of 3 and 5. But when we simply add these two sums together, we would be double-counting the numbers that are multiples of both 3 and 5 (i.e., the multiples of 15).

Thus, we need to subtract the sum of the multiples of 15 from our previous result. This can be done using the formula for the sum of an arithmetic series.

Before we dive into the calculations, we need to recall how to find the number of terms in a finite arithmetic series. We can use the following equation:

t = (last term – first term)/difference + 1

Here’s how it breaks down for each step:

1. Find the sum of the multiples of 3 below 1000.
First, we need to determine the largest multiple of 3 less than 1000. This would be 999 (which is 333 times 3). Therefore, repeatedly adding 3 gives an arithmetic series with 333 terms. The sum can be calculated using the formula:

Sum = (n/2) * (first term + last term)

Sum of multiples of 3 = (333/2) * (3 + 999) = 166,833

2. Find the sum of the multiples of 5 below 1000.
The largest multiple of 5 less than 1000 is 995 (which is 199 times 5). There are 199 terms in this series. The sum can be calculated as:

Sum of multiples of 5 = (199/2) * (5 + 995) = 99,500

3. Find the sum of the multiples of 15 (3 * 5) below 1000.
The largest multiple of 15 less than 1000 is 990 (which is 66 times 15). There are 66 terms in this series. The sum can be calculated as:

Sum of multiples of 15 = (66/2) * (15 + 990) = 33,165

Now, let’s add the sums of multiples of 3 and 5, and then subtract the sum of multiples of 15:

Sum of multiples of 3 or 5 = Sum (multiples of 3) + Sum (multiples of 5) – Sum (multiples of 15)

Put the values into the equation:

Sum of multiples of 3 or 5 = 166,833 + 99,500 – 33,165 = 233,168.

More Answers: