What will be the least number which when doubled will be exactly divisible by 6 7 24 30

Solution:

We will be using the concept of LCM(Least Common Multiple) to solve this.

To determine the least number which when divided by 6, 15, and 18 leave the remainder 5 in each case,we need to find the LCM of the three given numbers.

Since, the LCM obtained will be the smallest common multiple of all the three numbers 6, 15, and 18, after getting LCM we need to add 5  to it so as to get 5 as a remainder.

Let's find the LCM of 6, 5 and 18 as shown below.

Therefore, LCM of 6, 15 and 18 = 2 × 3 × 3 × 5 = 90.

Thus we can see that, 90 is the least number exactly divisible by 6, 15, and 18.

To get a remainder 5, we need to add 5 to the LCM.

⇒ 90 + 5 = 95.

Thus, when 95 is divided by 6, 15, and 18 we get a remainder of 5 in each case.

Hence, the required number for the given problem is 95.

You can also use the LCM Calculator to solve this.

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 Question 8

Summary:

The least number which when divided by 6, 15, and 18 leaving a remainder of 5 in each case will be 95.

☛ Related Questions:

Which is the least number which when double will be exactly divisible by 12, 18, 21 and 30? [A]196 [B]630 [C]1260 [D]2520

630 The LCM of 12, 18, 21, 30 is 1260. ∴ The required Number = $latex = \frac{1260}{2} = 630&s=1$

Hence option [B] is correct answer.