A school has between 150 and 200 students enrolled. Every afternoon, all the students come together to participate in gym class. The students are separated into six distinct sections of students. If one student is absent from school, the sections can all have the same number of students. What is the sum of all possible numbers of students enrolled at the school?
If there are $s$ students, then $s-1$ must be divisible by 6. In other words, we want to find the sum of all values of $s$ for which $s-1\equiv 0\pmod{6}$. The multiples of 6 in the given range are 150, 156, ..., 198, so the possible values of $s$ are 151, 157, ..., 199. Recalling that the sum of an arithmetic series is  \[
\frac{(\text{first term}+\text{last term})(\text{number of terms})}{2},
\]we find that these integers sum to $(151+199)(9)/2=\boxed{1575}$.
1575
Level 4
Number Theory
Problem: A school has between 150 and 200 students enrolled. Every afternoon, all the students come together to participate in gym class. The students are separated into six distinct sections of students. If one student is absent from school, the sections can all have the same number of students. What is the sum of all possible numbers of students enrolled at the school?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the total number of students, let the number of students be denoted as N. Given that N is between 150 and 200, we can express the conditions mathematically. The students are divided into six sections, so for the sections to have the same number of students when one student is absent, N - 1 must be divisible by 6. This means we can write the equation:

N - 1 ≡ 0 (mod 6)

This implies:

N ≡ 1 (mod 6)

Next, we need to identify the values of N that satisfy this condition within the range of 150 to 200. We can start by finding the smallest number greater than or equal to 150 that satisfies N ≡ 1 (mod 6). 

Calculating the first few values:
- 150 mod 6 = 0, so 150 does not satisfy the condition.
- 151 mod 6 = 1, so 151 satisfies the condition.
- 152 mod 6 = 4, so 152 does not satisfy the condition.
- 153 mod 6 = 3, so 153 does not satisfy the condition.
- 154 mod 6 = 2, so 154 does not satisfy the condition.
- 155 mod 6 = 5, so 155 does not satisfy the condition.
- 156 mod 6 = 0, so 156 does not satisfy the condition.
- 157 mod 6 = 1, so 157 satisfies the condition.
- Continuing this process, we find that the numbers satisfying N ≡ 1 (mod 6) in the range are: 151, 157, 163, 169, 175, 181, 187, 193, and 199.

Next, we sum these valid values:
151 + 157 + 163 + 169 + 175 + 181 + 187 + 193 + 199 = 1575.

Thus, the sum of all possible numbers of students enrolled at the school is 1575.<|END_OF_LATENT|> The final answer is $1575$.