Question

What is the smallest number which when decreased by 8 is divisible by 21, 27, 33, and 55?.

Answer 1

The required number which when decreased by 8 is divisible by 21, 27, 33, and 55 is 10403.

We need to find the smallest number which when decreased by 8 is divisible by 21, 27, 33, and 55.

What is LCM?

The least common multiple is also known as LCM (or) the lowest common multiple in math. The least common multiple of two or more numbers is the smallest number among all common multiples of the given numbers.

We find the LCM of 21, 27, 33, and 55.

3 | 21   27   33   55

3 | 7     9     11    55

3 | 7     3     11    55

7 | 7     1     11    55

5 | 1     1     11    55

11 | 1     1     11    11

 | 1     1     1     1

When decreased by 8, we have to add 8 in the LCM of the numbers.

That is 10395+8=10403

Therefore, the required number when decreased by 8 is divisible by 21, 27, 33, and 55 is 10403.

To learn more about the LCM of numbers visit:

brainly.com/question/10942748.

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