What is the least four digit number which when diminished by 10 is divisible by 10, 20, 30 and 36?

Answer

Hint: In this question, we need to find the smallest number which when diminished by \[3\] is divisible by \[14,28,36,45\] . First, we need to find the factors of all the given numbers one by one by using the prime factorization method. Then we need to multiply the factors with the highest degree to find the least common multiple of the given numbers. Then we will find the required answer by adding \[3\] to the obtained LCM where LCM is the Least Common multiple is defined as the number divisible by all the numbers.

Complete step-by-step answer:

So, the correct answer is “Option C”.

Note: In order to solve these types of questions, we should have a strong grip over tests for Divisibility of Numbers and prime factorization. We also need to know that the Prime factorization is a method of factoring a given number such that the obtained factors contain only prime numbers and the factors are known as prime factors. We also need to know that here, given that the smallest number is diminished, that we need to subtract by a number, then the number which is divisible has to be added with the same number.

Solution:

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

We know that the smallest 4-digit number is 1000.

Hence,the LCM of 18, 24 and 32 is calculated as shown below,

Therefore, LCM of 18, 24 and 32 = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

Thus, we have 288 as the smallest number, which is exactly divisible by 18, 24, and 32.

Since it's not a 4-digit number, we need to find the multiple of 288, close to 1000.

Here we have 136 as the remainder.

Therefore, we need to subtract 136 and add 288 to make the smallest 4 digit number exactly divisible by18, 24, and 32.

So, the multiple of 288 just above 1000 is: 1000 – 136 + 288 = 1152.

Hence, the smallest 4-digit number which is divisible by 18, 24, and 32 is 1152.

You can also use the LCM Calculator to solve this.

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 Question 9

Summary:

The smallest 4-digit number which is exactly divisible by 18,24, and 32, is 1152.

☛ Related Questions:

As per the norms of the government, the UPSC has released the Public Disclosure of Marks for the non-recommended willing candidates on the official website of the commission. A PDF has been released which contains the marks and other details of the non-recommended willing candidates. Earlier, the UPSC Examinations 2021 Reserved List was released. 63 candidates have been recommended by the commission to fill in the remaining posts. With reference to the 2022 exam cycle, The Union Public Service Commission (UPSC) examination was conducted on the 16th, 17th, 18th, 24th, and 25th of September 2022. This is one of the most coveted jobs in India. The candidates are required to go through a 3 stage selection process - Prelims, Main and Interview. The marks of the main examination and interview will be taken into consideration while preparing the final merit list. The candidates must go through the UPSC Civil Service mains strategy to have an edge over others.