The larger of two consecutive odd integers is three times the smaller.  What is their sum?
Let the smaller of the integers be $x$.  Then the larger is $x + 2$. So $x + 2 = 3x$, from which $x = 1$.  Thus the two integers are 1 and 3, and their sum is $\boxed{4}$.
4
Level 4
Prealgebra
Problem: The larger of two consecutive odd integers is three times the smaller.  What is their sum?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>Let the smaller odd integer be represented as x. The next consecutive odd integer can then be expressed as x + 2. According to the problem, the larger integer (x + 2) is equal to three times the smaller integer (x). This can be formulated as the equation:

x + 2 = 3x.

To solve for x, first, isolate x on one side of the equation. Subtract x from both sides:

2 = 3x - x,
2 = 2x.

Next, divide both sides by 2 to find the value of x:

x = 1.

Now that we have the smaller integer, we can find the larger integer by substituting x back into the expression for the larger integer:

Larger integer = x + 2 = 1 + 2 = 3.

To find the sum of the two integers, we add the smaller and larger integers together:

Sum = x + (x + 2) = 1 + 3 = 4.

Thus, the final answer is 4.<|END_OF_LATENT|> The final answer is $4$.