In a set of five consecutive integers, the largest integer is less than twice the average of the five integers. What is the smallest integer that could be in the set?
Let the integers be $n$, $n+1$, $n+2$, $n+3$ and $n+4$.  Their average is $n+2$, so we have $$n+4<2(n+2) \Rightarrow n+4<2n+4 \Rightarrow 0<n.$$Thus $n$ is at least $\boxed{1}$.
1
Level 4
Prealgebra
Problem: In a set of five consecutive integers, the largest integer is less than twice the average of the five integers. What is the smallest integer that could be in the set?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, let the five consecutive integers be represented as n, n+1, n+2, n+3, and n+4. The largest integer in this set is n+4. 

Next, we calculate the average of these five integers. The sum of the integers is n + (n+1) + (n+2) + (n+3) + (n+4) = 5n + 10. The average is then (5n + 10) / 5 = n + 2.

According to the problem, the largest integer (n + 4) must be less than twice the average. Therefore, we set up the inequality:

n + 4 < 2(n + 2).

Now, we simplify the right side of the inequality:

n + 4 < 2n + 4.

Next, we isolate n by subtracting n and 4 from both sides:

0 < n.

This tells us that n must be greater than 0. The smallest integer satisfying this condition is n = 1.

Thus, the smallest integer in the set of five consecutive integers is 1.<|END_OF_LATENT|> The final answer is $1$.