Question

find the smallest natural number which when divide 8 10 and 12 leaves the remainder 2 in each case​

Answer 1

The required number which when divide 8, 10 and 12 leaves the remainder 2 in each case is 122.

Step-by-step explanation:

To find : The smallest natural number which when divide 8, 10 and 12 leaves the remainder 2 in each case​?  

Solution :  

We find the LCM of 8, 10 and 12.

2 | 8  10  12

2 | 4   5   6

2 | 2   5   3

3 | 1    5   3

5 | 1    5   1

  | 1     1   1

$LCM(8,10,12)=2\times 2\times 2\times 3\times 5$  

$LCM(8,10,12)=120$  

In each case it leaves a remainder 2 so we add 2 in the LCM of the numbers.

i.e. 120+2=122.

Therefore, the required number which when divide 8, 10 and 12 leaves the remainder 2 in each case is 122.

#Learn more

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