# Shobute🍻

## Project-Euler / 1

Find the sum of all the multiples of 3 or 5 below 1000.

Very simple, just go through every number between 0 and 1000, and if it’s a multiple of 3 or 5 add it to the running total.

Here’s my initial solution in Java:

public class Problem1 {
public static void main(String[] args) {
int sum = 0;
for (int i = 0; i < 1000; i++) {
if (i % 3 == 0 || i % 5 == 0) sum += i;
}
System.out.println(sum);
}
}

This is fine if you only want to find the multiples below 1000, but for big ranges we can do far better.

We can solve this problem in constant time by making use of the fact that: $$\sum_{i=1}^ni = \frac{n(n+1)}2$$

The sum of all multiples of 3 and 5 below n can be calculated as: $$3\sum_{i=1}^{\lfloor\tfrac{1}{3}n-1\rfloor}i + 5\sum_{i=1}^{\lfloor\tfrac{1}{5}n-1\rfloor}i - 15\sum_{i=1}^{\lfloor\tfrac{1}{15}n-1\rfloor}i$$

Multiples of 15 have to be subtracted to remove duplicates which were multiples of both 5 and 3. One needs to be subtracted from each range because the questions says below 1000.

Using the earlier formula, this expands to: $$\frac{3\lfloor\tfrac{1}{3}m\rfloor(\lfloor\tfrac{1}{3}m\rfloor+1) + 5\lfloor\tfrac{1}{5}m\rfloor(\lfloor\tfrac{1}{5}m\rfloor+1) - 15\lfloor\tfrac{1}{15}m\rfloor(\lfloor\tfrac{1}{15}m\rfloor+1)}2$$ where $$m = n -1$$.

Translating into Java we get:

public class Problem1 {
public static void main(String[] args) {
final long RANGE = 1000;
long m = RANGE - 1;
long sum = (3 * (m / 3) * (m / 3 + 1) + 5 * (m / 5) * (m / 5 + 1) -
15 * (m / 15) * (m / 15 + 1)) / 2;
System.out.println(sum);
}
}

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