Answer :

Final answer:

To find the smallest number of 5 digits which is exactly divisible by 16, 25, 30, and 40, we need to find the smallest common multiple of these numbers. The prime factors of each number are 16 = 2 * 2 * 2 * 2, 25 = 5 * 5, 30 = 2 * 3 * 5, and 40 = 2 * 2 * 2 * 5. The highest powers of these prime factors give us the smallest number, which is 1200.

Explanation:

To find the smallest number of 5 digits which is exactly divisible by 16, 25, 30, and 40, we need to find the smallest common multiple of these numbers.

We can start by finding the prime factors of each number:

  • 16 = 2 * 2 * 2 * 2
  • 25 = 5 * 5
  • 30 = 2 * 3 * 5
  • 40 = 2 * 2 * 2 * 5

To find the smallest common multiple, we need to take the highest power of each prime factor:

  • Highest power of 2 = 2 * 2 * 2 * 2 = 16
  • Highest power of 3 = 3
  • Highest power of 5 = 5 * 5 = 25

Multiplying these highest powers gives us the smallest number which is exactly divisible by 16, 25, 30, and 40:

16 * 3 * 25 = 1200

So, the smallest number of 5 digits which is exactly divisible by 16, 25, 30, and 40 is 1200.

Learn more about Finding the smallest common multiple here:

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