Ball Distribution Problem Analysis

The problem describes a scenario involving distributing balls among students in a class. We are given information about the initial distribution and asked to find the remainder in a different distribution scenario.

Initial Ball Distribution

We know that 100 balls were distributed equally among students in a class, and 7 balls remained undistributed. Let the number of students in the class be $n$. The number of balls distributed equally is $100 - 7 = 93$. This means that 93 is perfectly divisible by the number of students, $n$.

Using the division algorithm, we can write this relationship as:

$\text{Total balls} = \text{Quotient} \times \text{Number of students} + \text{Remainder}$

$100 = q \times n + 7$

Where $q$ is the number of balls each student received, and $n$ is the number of students.

Subtracting 7 from both sides, we get:

$100 - 7 = q \times n$

$93 = q \times n$

This shows that $n$ must be a divisor of 93. We also know that the class strength ($n$) is between 10 and 60 (i.e., $10 < n < 60$).

Finding the Number of Students

Let's find the divisors of 93. The prime factorization of 93 is $3 \times 31$. The divisors of 93 are 1, 3, 31, and 93.

We need to find a divisor of 93 that falls within the range of 10 to 60. Let's check the divisors:

• 1: Not between 10 and 60.
• 3: Not between 10 and 60.
• 31: This is between 10 and 60 ($10 < 31 < 60$).
• 93: Not between 10 and 60.

The only divisor of 93 that is between 10 and 60 is 31. Therefore, the number of students in the class is 31.

Second Ball Distribution Scenario

Now, we are asked what would happen if there were 125 balls to begin with, distributed among the same number of students. We found that the number of students is 31.

We need to find the remainder when 125 balls are distributed among 31 students. We perform the division $125 \div 31$.

Let's divide 125 by 31:

$125 = \text{Quotient} \times 31 + \text{Remainder}$

We can find the largest multiple of 31 that is less than or equal to 125.

• $31 \times 1 = 31$
• $31 \times 2 = 62$
• $31 \times 3 = 93$
• $31 \times 4 = 124$
• $31 \times 5 = 155$ (Too large)

The largest multiple of 31 less than or equal to 125 is 124 ($31 \times 4$).

So, we can write:

$125 = 4 \times 31 + \text{Remainder}$

To find the remainder, we calculate $125 - (4 \times 31) = 125 - 124 = 1$.

The remainder when 125 balls are distributed among 31 students is 1.

Conclusion on Remaining Balls

When 125 balls are distributed among the students in the class (which has 31 students), 1 ball would remain undistributed.