Question

Find three consecutive odd integers whose sum is 117 n+ n+3 n+6=117

Answer 1

Let's denote the first odd integer as \( n \). Then, the three consecutive odd integers can be expressed as:

- First integer: \( n \)
- Second integer: \( n + 2 \)
- Third integer: \( n + 4 \)

However, you originally stated them as \( n \), \( n + 3 \), and \( n + 6 \). To keep them consecutive and consistent, let’s analyze using the first approach with odd integers:
- \( n \): first odd integer
- \( n + 2 \): second odd integer
- \( n + 4 \): third odd integer

However, if we want to stick with your original sequence:
- \( n \): first odd integer
- \( n + 3 \): second odd integer
- \( n + 6 \): third odd integer (which wouldn't be consecutive)

If we go by consecutive choices as initially mentioned, the correct representation for your question should rather maintain odd terms separated by 2. Thus we would come back to:

\[ n + (n + 2) + (n + 4) = 117 \]

This simplifies as:

\[
n + (n + 2) + (n + 4) = 117
\]
\[
3n + 6 = 117
\]

Now, subtract 6 from both sides:

\[
3n = 111
\]

Now, divide by 3:

\[
n = 37
\]

Now that we have the first integer:

- First odd integer: \( 37 \)
- Second odd integer: \( 39 \) (which is \( 37 + 2 \))
- Third odd integer: \( 41 \) (which is \( 39 + 2 \))

So, the three consecutive odd integers are \( 37, 39, \) and \( 41 \).

To verify:
\[
37 + 39 + 41 = 117
\]

Thus, the three consecutive odd integers whose sum is 117 are **37, 39, and 41**.