During the Inter - School Debate championship, students of the `9^(th)` grade of lIIions Public School had to be divided into group.It was found that if they were divided into group of four, one students was left out. If they were divided into group of six, then too, one students was left out. What was the minimum number of students in the grade such that they can be perfectly divided in group of five ?

Options

A. 15
B. 20
C. 25
D. 30

Answer

To solve the problem, we need to find the minimum number of students \( n \) in the 9th grade such that:

1. When divided into groups of 4, 1 student is left out.
2. When divided into groups of 6, 1 student is left out.
3. \( n \) is perfectly divisible by 5.

### Step-by-Step Solution:

1. **Understanding the Conditions**:
- From the first condition, we can express \( n \) as:
\[ n \equiv 1 \mod 4 \]
- From the second condition:
\[ n \equiv 1 \mod 6 \]
- From the third condition, we need:
\[ n \equiv 0 \mod 5 \]

2. **Finding Common Values**:
- We can start by finding numbers that satisfy the first two conditions. Since both conditions state that \( n \) leaves a remainder of 1 when divided by 4 and 6, we can express this as:
\[ n = 4k + 1 \quad \text{for some integer } k \]
\[ n = 6m + 1 \quad \text{for some integer } m \]
- Setting these equal gives us:
\[ 4k + 1 = 6m + 1 \]
\[ 4k = 6m \]
\[ 2k = 3m \]
- This means \( k \) must be a multiple of 3, so let \( k = 3j \):
\[ n = 4(3j) + 1 = 12j + 1 \]

3. **Finding Values of \( n \)**:
- Now, we can express \( n \) as:
\[ n = 12j + 1 \]
- We need \( n \) to also be divisible by 5:
\[ 12j + 1 \equiv 0 \mod 5 \]
- Simplifying \( 12 \mod 5 \):
\[ 12 \equiv 2 \mod 5 \]
- Thus, we have:
\[ 2j + 1 \equiv 0 \mod 5 \]
\[ 2j \equiv -1 \equiv 4 \mod 5 \]
- Multiplying both sides by the modular inverse of 2 modulo 5 (which is 3):
\[ j \equiv 3 \times 4 \mod 5 \]
\[ j \equiv 12 \mod 5 \equiv 2 \mod 5 \]
- Therefore, \( j \) can be expressed as:
\[ j = 5m + 2 \quad \text{for some integer } m \]

4. **Finding \( n \)**:
- Substituting back into the equation for \( n \):
\[ n = 12(5m + 2) + 1 = 60m + 24 + 1 = 60m + 25 \]
- The smallest value occurs when \( m = 0 \):
\[ n = 25 \]

5. **Verification**:
- Check if \( n = 25 \) satisfies all conditions:
- Divided by 4: \( 25 \div 4 = 6 \) remainder 1 (satisfied)
- Divided by 6: \( 25 \div 6 = 4 \) remainder 1 (satisfied)
- Divided by 5: \( 25 \div 5 = 5 \) remainder 0 (satisfied)

Thus, the minimum number of students in the grade is \( \boxed{25} \).