Question

Find the least number which when divided by 3 4 5 6 10 and 12 leaves a remainder of 2 in each case .

Answer 1

To find the least number which when divide by 5, 4 and 3 then leave remainder 2.

So, we have to calculate the L.C.M of the 5, 4, 3.

L.C.M of 5, 4 and 3 are 60.

Now, Just add 2 in 60.

So, 60 + 2 = 2.

62 is the least number which when divide by 5, 4 and 3 then leave remainder 2.

$\huge \red{thank \: you}$

Answer 2

The least number which when divided by 3, 4, 5, 6, 10, and 12 leaves a remainder of 2 in each case is 62.

Given: The numbers 3, 4, 5, 6, 10, and 12.

To Find: The least number which when divided by 3, 4, 5, 6, 10, and 12  leaves a remainder of 2 in each case.

Solution:

• We are required to find the least number, so we shall find the LCM of the given numbers.

• The LCM after being increased by a particular number gives us the lowest number which divides a set of numbers leaving a certain fixed remainder in each case.

Coming to the numerical, we have,

The numbers 3, 4, 5, 6, 10, and 12.

We need to find the least number so we shall find the LCM first using the prime factorization method.

Writing the prime factors of the numbers, we get;

3 = 1 × 3

4 = 2 × 2

5 = 1 × 5

6 = 2 × 3

10 = 2 × 5

12 = 2 × 2 × 3

∴ LCM ( 3, 4, 5, 6, 10, 12 ) = 2 × 2 × 3 × 5

                                            = 60

Now, it is said that the LCM must leave a remainder of 2 when it divides the numbers, so we shall increase the LCM by 2 so that the remainder of 2 stays always.

So, the increased number is = 60 + 2

                                               = 62

Hence, the least number when divided by 3, 4, 5, 6, 10, and 12 leaves a remainder of 2 in each case is 62.

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