Question

What is the least number which whe divided by either 2,4,6,8,10 or 12 leaves a remainder of 1 in each case?

Answer 1

required number is LCM of given numbers + 1
so LCM of 2,4,6,8,10 :
4= 2^2
6=2*3
8=2^3
10=2*5
LCM is 2^3 *3*5= 120
required number is 120+1= 121

Answer 2

Answer:

The least number  which when divided by either 2,4,6,8,10 or 12 leaves a remainder of 1 in each case is 121.          

Step-by-step explanation:

To find : What is the least number which when divided by either 2,4,6,8,10 or 12 leaves a remainder of 1 in each case?

Solution :

First we find the LCM of the numbers 2,4,6,8,10 or 12

2 | 2  4  6  8  10  12

2 | 1   2  3  4   5   6

2 | 1   1   3  2   5   3

3 | 1   1   3  1   5   3

5 | 1   1   1   1   5   1

  | 1   1   1   1   1   1

The LCM of the numbers is

$LCM=2\times 2\times 2\times 3\times 5$

$LCM=120$

The required number is $120+1=121$

Therefore, The least number  which when divided by either 2,4,6,8,10 or 12 leaves a remainder of 1 in each case is 121.