To solve this problem, we need three integers that meet two conditions: a range of 10 and a mean of 13.

Step 1: Understanding the range

The range is defined as the difference between the largest and smallest number. So, if the range is 10, then:

$\text{Largest integer} - \text{Smallest integer} = 10$

Let the three integers be $x_1$, $x_2$, and $x_3$, arranged in increasing order, i.e., $x_1 \leq x_2 \leq x_3$. Thus,

$x_3 - x_1 = 10$

Step 2: Understanding the mean

The mean is the average of the three integers. So, if their mean is 13, then:

$\frac{x_1 + x_2 + x_3}{3} = 13$

Multiplying both sides by 3:

$x_1 + x_2 + x_3 = 39$

Step 3: Solving the system of equations

We now have two key pieces of information:

1. $x_3 - x_1 = 10$
2. $x_1 + x_2 + x_3 = 39$

From equation 1, we can express $x_3$ in terms of $x_1$:

$x_3 = x_1 + 10$

Substitute this into the second equation:

$x_1 + x_2 + (x_1 + 10) = 39$

Simplifying:

$2x_1 + x_2 + 10 = 39$

$2x_1 + x_2 = 29$

Step 4: Finding suitable integers

Now, we need to find integer values of $x_1$, $x_2$, and $x_3$, all less than 25.

• Let's try $x_1 = 9$. Then: $2(9) + x_2 = 29 \implies 18 + x_2 = 29 \implies x_2 = 11$ Using $x_3 = x_1 + 10 = 9 + 10 = 19$.

So the three integers are $9, 11,$ and $19$.

Verification

• Range: $19 - 9 = 10$ (Correct)
• Mean: $\frac{9 + 11 + 19}{3} = \frac{39}{3} = 13$ (Correct)

Thus, the three integers are: $\boxed{9, 11, 19}$

Would you like further details or have any questions?

Related Questions:

1. Can you find another set of three integers with a mean of 13 and a different range?
2. How would the numbers change if the range was increased to 12 but the mean remained 13?
3. What would the mean be if the range stayed the same, but the largest number was 23?
4. Can you write an equation for the sum of four integers with a mean of 15?
5. How does changing the smallest integer affect the mean and range in general?

Tip:

When dealing with mean and range problems, setting up simple equations based on their definitions is a quick way to solve the problem efficiently.