To find the natural numbers between 101 and 999 that are divisible by both 2 and 5, we need numbers that are divisible by $2 \times 5 = 10$, since both conditions (divisibility by 2 and 5) are met by divisibility by 10.

Step-by-Step Solution

1. Define the range: We need numbers between 101 and 999.

2. Identify the first number divisible by 10 within the range:

   • The smallest number greater than or equal to 101 that is divisible by 10 is 110.

3. Identify the last number divisible by 10 within the range:

   • The largest number less than or equal to 999 that is divisible by 10 is 990.

4. Set up an arithmetic sequence:

   • The numbers divisible by 10 between 110 and 990 form an arithmetic sequence with the first term $a = 110$, the last term $l = 990$, and a common difference $d = 10$.

5. Calculate the number of terms in the sequence: $n = \frac{l - a}{d} + 1$ Substituting the values: $n = \frac{990 - 110}{10} + 1 = \frac{880}{10} + 1 = 88 + 1 = 89$

Conclusion

There are 89 natural numbers between 101 and 999 that are divisible by both 2 and 5.

Would you like more details, or do you have any questions?

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Here are some related questions you might find helpful:

1. What are the numbers between 101 and 999 that are divisible by 3 and 5?
2. How would we find numbers in a given range divisible by multiple factors?
3. How does the sequence approach work for counting divisible numbers?
4. How many numbers between 200 and 500 are divisible by both 2 and 7?
5. What is the smallest common multiple of two numbers?

Tip: When searching for numbers divisible by two factors, use their least common multiple (LCM) to simplify counting.