The mean (average) of 5 consecutive integers is 9. The smallest of these 5 integers is

1. Break Down the Problem

To find the smallest of 5 consecutive integers with a mean of 9, we start by letting the smallest integer be represented by $x$. The five consecutive integers can then be expressed as:

• $x$
• $x + 1$
• $x + 2$
• $x + 3$
• $x + 4$

2. Relevant Concepts

The mean (average) of these integers can be computed as follows: $\text{Mean} = \frac{\text{Sum of the integers}}{\text{Number of integers}} = \frac{x + (x + 1) + (x + 2) + (x + 3) + (x + 4)}{5}$

3. Analysis and Detail

First, we simplify the sum of the integers: $\text{Sum} = x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 5x + 10$

Next, we substitute this sum into the mean formula: $9 = \frac{5x + 10}{5}$

Multiplying both sides by 5 gives: $45 = 5x + 10$

Now, we isolate $x$: $5x = 45 - 10$ $5x = 35$ $x = 7$

4. Verify and Summarize

To verify, we find the five consecutive integers:

• $7, 8, 9, 10, 11$

Calculating their mean: $\text{Mean} = \frac{7 + 8 + 9 + 10 + 11}{5} = \frac{45}{5} = 9$ The calculations confirm that our assumption is correct.

Final Answer

The smallest of these 5 integers is $\boxed{7}$.