Find the smallest number which when diminished by 17 is divisible by each of 32, 36, 42 and 56

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:

Answer

Verified

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:

Here we need to find the smallest number which when diminished by \[3\]Is divisible by \[14,28,36,45\] .First let us find the factors of all the given numbers.Factors of \[14\] ,\[\Rightarrow \ 14 = 2 \times 7\]Factors of \[28\],\[\Rightarrow \ 28 = 2 \times 2 \times 7\]Factors of \[36\] ,\[\Rightarrow \ 36 = 2 \times 2 \times 3 \times 3\]Factors of \[45\] ,\[\Rightarrow \ 45 = 3 \times 3 \times 5\]Now let us find the LCM of \[(14,28,36,45)\] ,\[\Rightarrow \ 2 \times 2 \times 7 \times 3 \times 3 \times 5\]On multiplying,We get ,Thus the LCM is \[1260\]We are given that the smallest number is diminished by \[3\], to get divisible by all the numbers.The required number minus \[3\] gives the smallest number which is the LCM .That is \[\text{required number} – 3 = \text{LCM}\]\[\Rightarrow \ \text{required number} – 3 = 1260\]On adding both sides by \[3\] ,We get,\[\Rightarrow \ \text{required number} = 1260 + 3\]Thus our required number is \[1263\].Final answer :The smallest number which when diminished by \[3\] is divisible by \[14,28,36,45\] is \[1263\] .

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.

Answer

Verified

Hint:Here, we have to find the smallest number that is divisible by all the numbers. We will find the factors of all the given numbers by using the prime factorization method. Then we will 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 7 to the obtained LCM. Least Common multiple is defined as the number divisible by all the numbers.

Complete step by step solution:

We will find the smallest numbers that are divisible by all the numbers 12, 16, 18, 21 and 28.Now, we will find the LCM of these numbers We will find the factors for the numbers to find the LCM of these numbers by using the prime factorization method.We will find the factors of 12 by using the method of prime factorization.$\begin{array}{*{20}{l}} 2| {12} \\ \hline 2| 6 \\ \hline 3| 3 \\ \hline {}| 1 \end{array}$ Thus the factors of 12 are ${2^2} \times {3^1}$ We will find the factors of 16 by using the method of prime factorization.$\begin{array}{*{20}{l}} 2| {16} \\ \hline 2| 8 \\ \hline 2| 4 \\ \hline 2| 2 \\ \hline {}| 1 \end{array}$ Thus the factors of 16 are ${2^4}$ .We will find the factors of 18 by using the method of prime factorization.$\begin{array}{*{20}{l}} 2| {18} \\ \hline 3| 9 \\ \hline 3| 3 \\ \hline {}| 1 \end{array}$ Thus the factors of 18 are ${2^1} \times {3^2}$ We will find the factors of 21 by using the method of prime factorization.$\begin{array}{*{20}{l}} 3| {21} \\ \hline 7| 7 \\ \hline {}| 1 \end{array}$ Thus the factors of 21 are ${3^1} \times {7^1}$ We will find the factors of 28 by using the method of prime factorization. $\begin{array}{*{20}{l}} 2| {28} \\ \hline 2| {14} \\ \hline 7| 7 \\ \hline {}| 1 \end{array}$ Thus the factors of 28 are ${2^2} \times {7^1}$ Now, we will find the LCM of these factors.The LCM of these factors would be the highest exponent of the prime factors.LCM of 12, 16, 18, 21, 28 $ = {2^4} \times {3^2} \times {7^1}$ Multiplying the terms, we get$ \Rightarrow $ LCM of 12, 16, 18, 21, 28 $ = 1008$ We are given that the smallest number is diminished by 7, to get divisible by all the numbers.Required Number $ - 7 = $ Smallest divisible number$ \Rightarrow $ Required Number $ - 7 = $ LCM of all the numbers.$ \Rightarrow $ Required Number $ - 7 = $ 1008Adding 7 to both the sides, we get$ \Rightarrow $ Required Number $ = 1008 + 7$ $ \Rightarrow $ Required Number $ = 1015$ Therefore, the smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is 1015.

Thus, option (B) is the correct answer.

Note:

Here, we have used prime factorization to find prime factors of the given numbers. Prime factorization is a method of factoring a given number such that the obtained factors contain only prime numbers. These factors are known as prime factors. Since the smallest number is diminished i.e., subtracted by a number, then the number which is divisible has to be added with the same number.