Source: 2018 Canadian Open Math Challenge Part A Problem 2
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Let $v$, $w$, $x$, $y$, and $z$ be five distinct integers such that $45 = v\times w\times x\times y\times z.$ What is the sum of the integers?
1. We start with the given equation:
   \[
   45 = v \times w \times x \times y \times z
   \]
   where \(v, w, x, y, z\) are five distinct integers.

2. To find the sum of these integers, we need to factorize 45 into five distinct integers. The prime factorization of 45 is:
   \[
   45 = 3^2 \times 5 \times 1
   \]

3. Since the product of five distinct integers must equal 45, and some of these integers must be negative to ensure the product is positive, we consider the possible factorization:
   \[
   45 = 1 \times (-1) \times 3 \times (-3) \times 5
   \]

4. We verify that the product of these integers is indeed 45:
   \[
   1 \times (-1) \times 3 \times (-3) \times 5 = (1 \times -1) \times (3 \times -3) \times 5 = -1 \times -9 \times 5 = 9 \times 5 = 45
   \]

5. Now, we calculate the sum of these integers:
   \[
   1 + (-1) + 3 + (-3) + 5 = 1 - 1 + 3 - 3 + 5 = 0 + 0 + 5 = 5
   \]

Conclusion:
\[
\boxed{5}
\]