Final answer:
The smallest number greater than 100,000 that is exactly divisible by 8, 15, and 21 is 100,800. This number is found by calculating the least common multiple (LCM) of these numbers and determining the smallest multiple of the LCM that exceeds 100,000.
Step-by-step explanation:
To find a number greater than 100,000 that is exactly divisible by 8, 15, and 21, we need to determine the least common multiple (LCM) of these numbers. The LCM of 8, 15, and 21 is the smallest number that all three numbers divide into without leaving a remainder. We can calculate the prime factorization of each number: 8 = 23, 15 = 3 × 5, and 21 = 3 × 7.
To get the LCM, take the highest powers of all prime factors involved: LCM = 23 × 3 × 5 × 7 = 840. Now we need to find the smallest multiple of 840 that is greater than 100,000. Dividing 100,000 by 840 gives us approximately 119.04, which means we need to find the next whole number. 119 × 840 = 99,960, which is not greater than 100,000. The next multiple, 120 × 840, equals 100,800 which is the number we are looking for that meets all the criteria.