We are asked to find three consecutive odd integers that have a sum of 63.

We can represent the three consecutive odd integers as:

• The first odd integer: $2n + 1$
• The second odd integer: $2n + 3$
• The third odd integer: $2n + 5$

Now, let's set up an equation where the sum of these three integers equals 63:

$(2n + 1) + (2n + 3) + (2n + 5) = 63$

Step 1: Simplify the equation

Combine like terms:

$2n + 2n + 2n + 1 + 3 + 5 = 63$

$6n + 9 = 63$

Step 2: Solve for $n$

Subtract 9 from both sides:

$6n = 63 - 9$

$6n = 54$

Now divide by 6:

$n = \frac{54}{6} = 9$

Step 3: Find the three consecutive odd integers

Now that we have $n = 9$, substitute it back into the expressions for the three integers:

• First integer: $2(9) + 1 = 18 + 1 = 19$
• Second integer: $2(9) + 3 = 18 + 3 = 21$
• Third integer: $2(9) + 5 = 18 + 5 = 23$

Conclusion:

The three consecutive odd integers are 19, 21, and 23.

To verify, their sum is:

$19 + 21 + 23 = 63$

Thus, the solution is correct.

Would you like any further clarification or details?

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Here are 5 related questions:

1. How would the equation change if the sum were 75 instead of 63?
2. What happens if we want