Question

Determine the number nearest 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.

Updated on May 31, 2025 | By Learnzy Admin | 👁️ Views: 371 students

Solution
✔ Verified

Step 1: Find the Least Common Multiple (LCM) of 8, 15, and 21.

Prime factorization:
8 = 2³
15 = 3 × 5
21 = 3 × 7
LCM = 2³ × 3 × 5 × 7 = 8 × 3 × 5 × 7 = 840

So, any number divisible by 8, 15, and 21 must be a multiple of 840.

Step 2: Find the multiple of 840 that is nearest to 110000 and greater than 100000.

Divide 110000 by 840:
110000 ÷ 840 ≈ 130.95
Now check nearby multiples:
130 × 840 = 109200
131 × 840 = 110040
132 × 840 = 110880

Among these, 110040 is the nearest to 110000 and is greater than 100000.

Hence answer is 110040

Was this solution helpful? 28
Click here to download practice questions on Real Numbers

More Questions on Real Numbers

Question 1

In a school there are two sections – section A and section B of class X. There are 32 students in section A and 36 students in section B. Determine the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B.


View solution
Question 2

If HCF(6, a) = 2 and LCM(6, a) = 60 then find the value of a.


View solution
Question 3

Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.


View solution
Question 4

Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.


View solution
Question 5

Prove that 7×11×13 +13 is a composite number.


View solution
Question 6

If the HCF of two numbers is 16 and their product is 3072, find their LCM.


View solution
Question 7

Find the smallest number that when divided by 12, 16, and 24 leaves a remainder of 4 in each case.


View solution
Question 8

Find the smallest number that is divisible by 6, 10, and 15, and leaves a remainder of 5 in each case.


View solution
Question 9

The product of HCF and LCM of 60,84 and 108 is


View solution
Question 10

The LCM of 60, 84 and 108 is


View solution
Question 11

If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32 , 36) is


View solution