A number is divisible by 5 and 6. It may not be divisible by

(a) 10
B). 15
C). 30
D) 60

To solve this problem, we need to analyze the divisibility properties of a number that is divisible by both 5 and 6.

### Key Insights:
1. **Divisibility by 5 and 6** implies that the number is divisible by the **Least Common Multiple (LCM)** of 5 and 6.
   - LCM(5, 6) = \(30\). Hence, the number must be divisible by 30.

2. For divisibility by the given options, let's check:
   - **(a) 10**: A number divisible by 30 is divisible by 10, as 30 is a multiple of 10.
   - **(b) 15**: A number divisible by 30 is divisible by 15, as 30 is a multiple of 15.
   - **(c) 30**: By definition, the number is divisible by 30.
   - **(d) 60**: A number divisible by 30 **may or may not** be divisible by 60, depending on whether it contains an additional factor of 2 (since \(60 = 30 \times 2\)).

### Conclusion:
The correct answer is **(D) 60**, as the number may not be divisible by 60.

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Would you like further explanation or help with related concepts? Here are five questions for exploration:

1. What is the LCM, and how is it calculated?
2. Why is divisibility by a number related to its factors?
3. How does the presence of additional factors (like 2 in 60) change divisibility rules?
4. What are some real-life applications of divisibility and LCM?
5. How can prime factorization help with divisibility questions?

**Tip**: When solving divisibility problems, always consider the LCM to simplify the calculations and reasoning.

A number is divisible by 5 and 6. It may not be divisible by (a) 10, (b) 15, (c) 30, (d) 60.

Explore the divisibility rules of numbers that are divisible by both 5 and 6. This problem shows how to determine whether such a number is divisible by 10, 15, 30, or 60. Learn about LCM, divisibility properties, and their practical applications. Suitable for students in Grades 6-8.

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