In a class, if when 1 student is absent the remaining students can be divided into 6 equal groups, and when 2 students are absent the remaining students can be divided into 7 equal groups, what is the total strength of the class?

Introduction / Context:
This arithmetic reasoning question checks your understanding of divisibility, remainders, and how to translate conditions about grouping students into a mathematical problem. The class strength must satisfy two divisibility conditions based on how many students are absent. By converting the statements into congruence relations, we can solve for the total number of students using simple modular arithmetic.

Given Data / Assumptions:
- Let the total number of students in the class be N.- When 1 student is absent, N - 1 is divisible by 6.- When 2 students are absent, N - 2 is divisible by 7.- We seek a positive integer N that satisfies both conditions simultaneously.

Concept / Approach:
The key idea is to model the conditions using congruences. The statement that N - 1 is divisible by 6 means N - 1 is a multiple of 6, or N is 1 more than a multiple of 6. In modular arithmetic, this is written as N ≡ 1 (mod 6). Similarly, N - 2 divisible by 7 means N ≡ 2 (mod 7). We then need a number that satisfies both congruences at the same time. Because the moduli 6 and 7 are relatively prime, we can use simple trial or the Chinese remainder idea on numbers within a reasonable range, or we can reason directly by listing candidates.

Step-by-Step Solution:
- Condition 1: N ≡ 1 (mod 6). So N can be 1, 7, 13, 19, 25, 31, 37, and so on, each obtained by adding 6.- Condition 2: N ≡ 2 (mod 7). So N can be 2, 9, 16, 23, 30, 37, 44, and so on, each obtained by adding 7.- Find the common value in both sequences.- Looking at the lists, the number 37 appears in both sequences.- Therefore N = 37 satisfies both conditions.- Check: N - 1 = 36, which is divisible by 6, and N - 2 = 35, which is divisible by 7.

Verification / Alternative check:
We can verify quickly by substituting the answer into the original statement. With 37 total students, if 1 is absent, there are 36 present, which can be divided into 6 equal groups of 6 students each. If 2 students are absent, there are 35 present, which can be divided into 7 equal groups of 5 students each. Both conditions are fully satisfied without contradiction, confirming that 37 is correct.

Why Other Options Are Wrong:
- Option 28: 28 - 1 = 27 does not divide evenly by 6, so the first condition fails.- Option 19: 19 - 1 = 18 works for divisibility by 6, but 19 - 2 = 17 is not divisible by 7.- Option 57: 57 - 1 = 56 is not a multiple of 6, so the first condition fails.- Option 49: 49 - 2 = 47 is not divisible by 7, so the second condition fails.

Common Pitfalls:
A common mistake is to misread the problem and treat 6 and 7 as the group sizes when students are present rather than understanding that these refer to equal grouping possibilities after some are absent. Another pitfall is to ignore one of the divisibility conditions, solving only for one congruence and not checking the other. Always translate such word problems into clear congruence statements and then systematically search for a value that satisfies all of them.

Final Answer:
The total number of students in the class is 37.