16.	In a collection of distinct positive integers, the following conditions apply: 
•	Three of the integers have a remainder of 0 when divided by 3.
•	Fifteen of the integers have a remainder of 0 when divided by 11. 
•	X is the maximum value in the collection. 
What is the smallest possible value of X?
a)	165
b)	187
c)	198
d)	33


To solve this problem, let's define the conditions step by step and find the smallest possible value of \( X \).

### Step 1: Identify the Conditions

1. **Three integers have a remainder of 0 when divided by 3.**
   - This means that three integers in the set are divisible by 3.
   - These integers are multiples of 3.

2. **Fifteen integers have a remainder of 0 when divided by 11.**
   - This means that fifteen integers in the set are divisible by 11.
   - These integers are multiples of 11.

3. \( X \) is the maximum value in the collection. We want to find the smallest possible value of \( X \).

### Step 2: Set the Problem in Terms of the Least Common Multiple (LCM)

To satisfy both conditions, the integers divisible by 3 and 11 should have overlap, because we want to minimize \( X \). So, we focus on multiples of both 3 and 11, which are given by the **LCM of 3 and 11**.

The least common multiple (LCM) of 3 and 11 is:

\[
\text{LCM}(3, 11) = 33
\]

### Step 3: Determine the Smallest Possible Value of \( X \)

Since we need 3 multiples of 3 and 15 multiples of 11, let's check the smallest value of \( X \) for the intersection of these two conditions:

- The smallest integer that satisfies both conditions (divisible by both 3 and 11) is \( 33 \), which is the LCM.
- The next multiples of 33 are \( 66, 99, 132, 165, 198, \ldots \).

Now, we are told that **three of the integers** are divisible by 3, and **fifteen of the integers** are divisible by 11. We can use the following integers that are divisible by both:

- \( 33, 66, 99, 132, 165, 198 \dots \)

Among these, **the smallest possible value of \( X \) that satisfies all conditions is \( 165 \)**, because it is the smallest integer that is a multiple of both 3 and 11 and meets the divisibility requirements (3 multiples of 3, 15 multiples of 11).

### Step 4: Verify with the Options

Looking at the answer choices:

a) 165  
b) 187  
c) 198  
d) 33

The smallest value that satisfies all the conditions is \( \boxed{165} \).

### Final Answer: \( \boxed{165} \)

---

Would you like further details or clarifications on any of the steps? 

Here are five related questions to consider:
1. How do we calculate the least common multiple (LCM) of two numbers?
2. What are other methods to approach divisibility problems like this?
3. What if the set had additional conditions, like divisibility by another number?
4. How do we handle overlapping conditions in integer sets mathematically?
5. Can we use prime factorization to solve problems like this more efficiently?

**Tip**: When dealing with divisibility problems involving two numbers, always start by checking the least common multiple (LCM) to minimize overlap and ensure you cover both conditions efficiently.

In a collection of distinct positive integers, the following conditions apply: • Three of the integers have a remainder of 0 when divided by 3. • Fifteen of the integers have a remainder of 0 when divided by 11. • X is the maximum value in the collection. What is the smallest possible value of X? a) 165 b) 187 c) 198 d) 33

Explore how to determine the smallest possible value of X in a set of distinct integers under specific divisibility conditions by 3 and 11. This problem focuses on the application of number theory and least common multiple (LCM) concepts, suitable for high school students in Grades 9-12.

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