 # The smallest number which when diminished by $3$ is divisible by $14,28,36,45$ isA. $1257$B. $1260$C. $1263$D. None of the above Verified
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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.

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 $1263$.
The smallest number which when diminished by $3$ is divisible by $14,28,36,45$ is $1263$ .