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**Hint:**First of all find the least common multiple of all the four numbers 27, 35, 25 and 21 given. Use the prime factorization method to find the L.C.M. First write the given numbers as the product of their prime factors one – by – one. Now, if a prime factor will be repeating then write them in exponential form. Take the product of all the different prime factors along with their highest exponent to get the L.C.M. Once the L.C.M is found, subtract 3 from it to get the answer.

**Complete step by step solution:**

Here we have been asked to find the smallest number in which we will add 3 so that it will be divisible by all the four numbers 27, 35, 25 and 21. First we need to find the L.C.M of these four numbers.

Now, the smallest number that will be divisible by each of the numbers 27, 35, 25 and 21 will be their L.C.M, so let us find the L.C.M of these numbers by the prime factorization method. So, writing all the numbers as the product of their prime factors one by one we get,

$\Rightarrow 27=3\times 3\times 3={{3}^{3}}$

$\Rightarrow 35=5\times 7$

$\Rightarrow 25=5\times 5={{5}^{2}}$

$\Rightarrow 21=3\times 7$

Therefore, the L.C.M will be the product of all the different prime factors with their highest exponent, so we get,

$\Rightarrow $ L.C.M = ${{3}^{3}}\times {{5}^{2}}\times 7$

$\Rightarrow $ L.C.M = 4725

Now, assuming the number in which we will add 3 so that it will be divisible by all the four numbers as x, we must have the mathematical condition given as: -

$\begin{align}

& \Rightarrow x+3=4725 \\

& \Rightarrow x=4725-3 \\

& \therefore x=4722 \\

\end{align}$

**So, the correct answer is “Option b”.**

**Note:**Note that if we are asked to calculate the highest number which can divide all the given four numbers then in that case we take the H.C.F of the given numbers. So you must not get confused while solving the question. Do not consider the obtained L.C.M as the answer because we need to subtract 3 from the L.C.M to get the answer as per the question.

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