In a nuclear power plant a technician is allowed an interval of maximum 100 minutes. A timer with a bell rings at specific intervals of time such that the minutes when the timer rings are not divisible by 2, 3, 5 and 7. The last alarm rings with a buzzer to give time for decontamination of the technician. How many times will the bell ring within these 100 minutes and what is the value of the last minute when the bell rings for the last time in a 100 minute shift?

Options

A. 25 times, 89
B. 21 times, 97
C. 22 times, 97
D. 19 times, 97

Answer

To solve the problem, we need to determine how many minutes within a 100-minute interval are not divisible by 2, 3, 5, and 7. We will also find the last minute when the timer rings.

### Step-by-Step Solution:

1. **Identify the Total Minutes**: We are considering the range from 1 to 100 minutes.

2. **Count Multiples of Each Number**:
- **Multiples of 2**: The multiples of 2 from 1 to 100 are 2, 4, 6, ..., 100. This forms an arithmetic sequence where:
- First term (a) = 2
- Common difference (d) = 2
- Last term (l) = 100
- Number of terms (n) = l/d = 100/2 = 50.
- **Multiples of 3**: The multiples of 3 from 1 to 100 are 3, 6, 9, ..., 99. This forms an arithmetic sequence:
- First term (a) = 3
- Common difference (d) = 3
- Last term (l) = 99
- Number of terms (n) = (99 - 3)/3 + 1 = 33.
- **Multiples of 5**: The multiples of 5 from 1 to 100 are 5, 10, 15, ..., 100. This forms an arithmetic sequence:
- First term (a) = 5
- Common difference (d) = 5
- Last term (l) = 100
- Number of terms (n) = 100/5 = 20.
- **Multiples of 7**: The multiples of 7 from 1 to 100 are 7, 14, 21, ..., 98. This forms an arithmetic sequence:
- First term (a) = 7
- Common difference (d) = 7
- Last term (l) = 98
- Number of terms (n) = 98/7 = 14.

3. **Apply Inclusion-Exclusion Principle**:
- We need to subtract the counts of numbers that are counted multiple times (i.e., those that are multiples of combinations of 2, 3, 5, and 7).
- Calculate the overlaps:
- **Multiples of 6 (2 and 3)**: 6, 12, ..., 96 → 16 terms.
- **Multiples of 10 (2 and 5)**: 10, 20, ..., 100 → 10 terms.
- **Multiples of 14 (2 and 7)**: 14, 28, ..., 98 → 7 terms.
- **Multiples of 15 (3 and 5)**: 15, 30, ..., 90 → 6 terms.
- **Multiples of 21 (3 and 7)**: 21, 42, ..., 84 → 4 terms.
- **Multiples of 35 (5 and 7)**: 35, 70 → 2 terms.
- **Multiples of 30 (2, 3, and 5)**: 30, 60, 90 → 3 terms.
- **Multiples of 42 (2, 3, and 7)**: 42 → 1 term.
- **Multiples of 70 (2, 5, and 7)**: 70 → 1 term.
- **Multiples of 105 (3, 5, and 7)**: None in this range.

4. **Final Calculation**:
- Total divisible by 2, 3, 5, or 7:
\[ = (50 + 33 + 20 + 14) - (16 + 10 + 7 + 6 + 4 + 2 + 3 + 1 + 1) = 117 - 50 = 67 \]
- Therefore, the numbers not divisible by 2, 3, 5, or 7:
\[ = 100 - 67 = 33 \]

5. **Finding the Last Minute**:
- We need to find the largest number less than or equal to 100 that is not divisible by 2, 3, 5, or 7.
- Checking from 100 downwards:
- 100 (divisible by 2)
- 99 (divisible by 3)
- 98 (divisible by 2)
- 97 (not divisible by 2, 3, 5, or 7)
- The last minute when the bell rings is **97**.

### Final Answer:
- The bell will ring **33 times** within 100 minutes.
- The last minute when the bell rings is **97**.