Find the lowest 6-digit number that is exactly divisible by 16, 25, 30, and 40.

The lowest 6-digit number that is exactly divisible by 16, 25, 30, and 40 is found by calculating the least common multiple of these numbers, which is 1200. The smallest 6-digit number that is a multiple of 1200 is 100800. Therefore, the answer is 100800.
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Answered by Anonymous | 2025-07-15

To find the lowest number of 6 digits that is exactly divisible by 16, 25, 30, and 40, we need to find the Least Common Multiple (LCM) of these numbers. Here's a step-by-step approach:

Factorize each number into its prime factors:

16 = 2^4
25 = 5^2
30 = 2 × 3 × 5
40 = 2^3 × 5


Determine the highest power of each prime number in these factorizations:

The highest power of 2 is 2^4 (from 16).
The highest power of 3 is 3 (from 30).
The highest power of 5 is 5^2 (from 25).


Calculate the LCM by multiplying these highest powers:
LCM = 2 4 × 3 1 × 5 2 = 16 × 3 × 25
Calculate step-by-step:

16 × 3 = 48
48 × 25 = 1200

Thus, the LCM of 16, 25, 30, and 40 is 1200.

Find the smallest 6-digit number and check for divisibility by this LCM:

The smallest 6-digit number is 100000.

To find the first 6-digit number divisible by 1200, divide 100000 by 1200 and round up to the nearest whole number:
1200 100000 ​ ≈ 83.3333

The smallest integer greater than 83.3333 is 84.



Now multiply this integer by the LCM to get the smallest 6-digit number divisible by all given numbers:
1200 × 84 = 100800


Therefore, the lowest 6-digit number that is exactly divisible by 16, 25, 30, and 40 is 100800.

Answered by EmmaGraceJohnson | 2025-07-22