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2652. Sum Multiples

Description

Given a positive integer n, find the sum of all integers in the range [1, n] inclusive that are divisible by 3, 5, or 7.

Return an integer denoting the sum of all numbers in the given range satisfying the constraint.

 

Example 1:

 Input: n = 7 Output: 21 Explanation: Numbers in the range [1, 7] that are divisible by 3, 5, or 7 are 3, 5, 6, 7. The sum of these numbers is 21.

Example 2:

 Input: n = 10 Output: 40 Explanation: Numbers in the range [1, 10] that are divisible by 3, 5, or 7 are 3, 5, 6, 7, 9, 10. The sum of these numbers is 40.

Example 3:

 Input: n = 9 Output: 30 Explanation: Numbers in the range [1, 9] that are divisible by 3, 5, or 7 are 3, 5, 6, 7, 9. The sum of these numbers is 30.

 

Constraints:

  • 1 <= n <= 103

Solutions

Solution 1: Enumeration

We directly enumerate every number $x$ in $[1,..n]$, and if $x$ is divisible by $3$, $5$, and $7$, we add $x$ to the answer.

After the enumeration, we return the answer.

The time complexity is $O(n)$, where $n$ is the given integer. The space complexity is $O(1)$.

Solution 2: Mathematics (Inclusion-Exclusion Principle)

We define a function $f(x)$ to represent the sum of numbers in $[1,..n]$ that are divisible by $x$. There are $m = \left\lfloor \frac{n}{x} \right\rfloor$ numbers that are divisible by $x$, which are $x$, $2x$, $3x$, $\cdots$, $mx$, forming an arithmetic sequence with the first term $x$, the last term $mx$, and the number of terms $m$. Therefore, $f(x) = \frac{(x + mx) \times m}{2}$.

According to the inclusion-exclusion principle, we can obtain the answer as:

\[f(3) + f(5) + f(7) - f(3 \times 5) - f(3 \times 7) - f(5 \times 7) + f(3 \times 5 \times 7)\]

The time complexity is $O(1)$, and the space complexity is $O(1)$.

  • class Solution { public int sumOfMultiples(int n) { int ans = 0; for (int x = 1; x <= n; ++x) { if (x % 3 == 0 || x % 5 == 0 || x % 7 == 0) { ans += x; } } return ans; } } 
  • class Solution { public: int sumOfMultiples(int n) { int ans = 0; for (int x = 1; x <= n; ++x) { if (x % 3 == 0 || x % 5 == 0 || x % 7 == 0) { ans += x; } } return ans; } }; 
  • class Solution: def sumOfMultiples(self, n: int) -> int: return sum(x for x in range(1, n + 1) if x % 3 == 0 or x % 5 == 0 or x % 7 == 0) 
  • func sumOfMultiples(n int) (ans int) { for x := 1; x <= n; x++ { if x%3 == 0 || x%5 == 0 || x%7 == 0 { ans += x } } return } 
  • function sumOfMultiples(n: number): number { let ans = 0; for (let x = 1; x <= n; ++x) { if (x % 3 === 0 || x % 5 === 0 || x % 7 === 0) { ans += x; } } return ans; } 
  • impl Solution { pub fn sum_of_multiples(n: i32) -> i32 { let mut ans = 0; for x in 1..=n { if x % 3 == 0 || x % 5 == 0 || x % 7 == 0 { ans += x; } } ans } }

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