The sum of three consecutive numbers is
45. What is the smallest of these
numbers?



Let the three consecutive numbers be \( x \), \( x + 1 \), and \( x + 2 \).

According to the problem, their sum is 45:
\[
x + (x + 1) + (x + 2) = 45
\]

Simplifying this equation:
\[
3x + 3 = 45
\]

Next, subtract 3 from both sides:
\[
3x = 42
\]

Finally, divide by 3:
\[
x = 14
\]

So, the smallest of these numbers is **14**.

Would you like more details or have any questions?

Here are five related questions:
1. What is the largest of the three consecutive numbers?
2. How would the equation change if the sum was 60 instead of 45?
3. What are the three consecutive numbers if the sum is 78?
4. How do you solve for four consecutive numbers with a given sum?
5. Can this problem be solved using algebraic expressions?

**Tip:** When dealing with consecutive numbers, express them using a single variable to simplify the equation.

The sum of three consecutive numbers is 45. What is the smallest of these numbers?

Learn how to solve for the smallest of three consecutive numbers when their sum is 45. This problem demonstrates key algebraic concepts, including forming and solving linear equations. Perfect for students in Grades 6-8.

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