Solution:

We will solve each part systematically.

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a) Integers divisible by 7 between 50 and 100

The formula for finding numbers divisible by $n$ in a range is: $n \cdot \lceil \frac{\text{start}}{n} \rceil, n \cdot \lceil \frac{\text{start}}{n} + 1 \rceil, \ldots, n \cdot \lfloor \frac{\text{end}}{n} \rfloor$

Here, $n = 7$, $\text{start} = 50$, $\text{end} = 100$.

1. Smallest multiple of $7 \geq 50$: $7 \cdot \lceil \frac{50}{7} \rceil = 7 \cdot 8 = 56$

2. Largest multiple of $7 \leq 100$: $7 \cdot \lfloor \frac{100}{7} \rfloor = 7 \cdot 14 = 98$

3. Sequence of multiples of $7$ between 50 and 100: $56, 63, 70, 77, 84, 91, 98$

4. Count of such numbers: $14 - 8 + 1 = 7$.

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b) Integers divisible by 11 between 50 and 100

Here, $n = 11$, $\text{start} = 50$, $\text{end} = 100$.

1. Smallest multiple of $11 \geq 50$: $11 \cdot \lceil \frac{50}{11} \rceil = 11 \cdot 5 = 55$

2. Largest multiple of $11 \leq 100$: $11 \cdot \lfloor \frac{100}{11} \rfloor = 11 \cdot 9 = 99$

3. Sequence of multiples of $11$ between 50 and 100: $55, 66, 77, 88, 99$

4. Count of such numbers: $9 - 5 + 1 = 5$.

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c) Integers divisible by both 7 and 11 between 50 and 100

For a number to be divisible by both $7$ and $11$, it must be divisible by their LCM: $\text{LCM}(7, 11) = 7 \cdot 11 = 77$

1. Smallest multiple of $77 \geq 50$: $77$.

2. Largest multiple of $77 \leq 100$: $77$.

3. Sequence of multiples of $77$ between 50 and 100: $77$

4. Count of such numbers: $1$.

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Final Answers:

a) $7$ integers: $56, 63, 70, 77, 84, 91, 98$
b) $5$ integers: $55, 66, 77, 88, 99$
c) $1$ integer: $77$

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Would you like a deeper explanation or visualization? 😊

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Related Questions:

1. How to generalize for any range and any divisor?
2. Can you explain how LCM is calculated step by step?
3. How would this change if the range were extended to negative integers?
4. What are the divisors of 77, and how do they relate to the problem?
5. How can modular arithmetic be used to find these solutions efficiently?

Tip:

When working with divisibility, understanding multiples and using the ceiling/floor functions can save significant calculation time.