Question

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

a.1490?

Answer 1

Answer:

In order to calculate smallest number divisible by 21,27,33,55 you first have to find LCM of these 4 no.

• LCM is = 10395
• Required Number = 10395 + 8 = 10403

Answer 2

Answer:

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

Step-by-step explanation:

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

Solution :  

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

$LCM(21,27,33,55)=3\times 3\times 3\times 7\times 5\times 11$  

$LCM(21,27,33,55)=10395$  

when decreased by 8,

So we add 8 in the LCM of the numbers.

i.e. 10395+8=10403.

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