What is the least number which when divided by 7 is divisible by each one of 21 28 36 and 45?

Answer

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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.

Given:

The number when divided by 21, 28, 36, and 45 and leaves the same remainder 4.

The number is also divisible by 13.

Calculations:

The least number which will be completely divisible by the 21, 28, 36, and 45 will be the LCM of 21, 28, 36, and 45

⇒ LCM(21, 28, 36, 45) = 1260

The least number will be completely divisible by the 21, 28, 36, and 45 and leaves remainder 4 = 1260k + 4

Where,

k = natural number

when k = 1

⇒ 1265

Here 1265 is not divisible by 13.

When k = 1, 2, 3, 4, 5. The number is not divisible by 13.

Now, k = 4

Number = 1260 × 4 + 4

⇒ Number = 5044

The number 5044 is divisible by 13