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What is the smallest number which when decreased by 8 is divisible by 21, 27, 33, and 55?a.1490?

Answer»

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.


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