To determine the number nearest to 100,000 but greater than 100,000 that is divisible by 8, 15, and 21, we follow these steps:

1. Find the Least Common Multiple (LCM) of 8, 15, and 21:

   • Prime factorization:
     • $8 = 2^3$
     • $15 = 3 \cdot 5$
     • $21 = 3 \cdot 7$
   • LCM is the product of the highest powers of all prime factors: $\text{LCM} = 2^3 \cdot 3 \cdot 5 \cdot 7 = 840$

2. Find the smallest multiple of 840 greater than 100,000:

   • Divide 100,000 by 840 and take the ceiling of the result: $\frac{100,000}{840} \approx 119.05$ The next integer is 120.
   • Multiply 120 by 840 to find the desired number: $120 \cdot 840 = 100,800$

Thus, the nearest number greater than 100,000 that is divisible by 8, 15, and 21 is 100,800.

Would you like a detailed breakdown of any step or further questions?

Related Questions:

1. How can I find the LCM of more than three numbers?
2. What are the properties of LCM and how is it different from GCD?
3. How would this process change for smaller or larger ranges?
4. Could this method work for numbers not mutually prime?
5. How can we verify that 100,800 is divisible by all three numbers?

Tip:

Always use the prime factorization method for finding the LCM to ensure accuracy, especially for larger numbers.