Question

Find the smallest number which, when increased by $5$ is exactly divisible by the number $110,135$ and $200$.

Answer 1

Step 1 Finding Least Common Multiple ( LCM ) of given numbers that are $\mathbf{110}\mathbf{,}\mathbf{}\mathbf{135}$and $\mathbf{200}$by prime factorisation method :

$$\begin{gathered} 110=2\times 5\times 11 \\ 135=3\times 3\times 3\times 5=3^3\times 5 \\ 200=2\times 2\times 2\times 5\times 5=2^3\times 5^2 \end{gathered}$$

Now Least Common Multiple of $110,135$and $200$ $=$Product of all the factors with highest powers

$$\begin{gathered} =2^3\times 3^3\times 5^2\times 11 \\ =8\times 27\times 25\times 11 \\ =59400 \end{gathered}$$

So Least Common Multiple of $110,135$ and $200$ $=$$59400$

Step 2 Finding the required smallest number by subtracting $\mathbf{5}$ from the Least Common Multiple obtained in Step 1

The smallest number which when increased by $5$ is exactly divisible by the number $110,135$ and $200$.$=$( Least Common Multiple of $110,135$ and $200$) $-5$

$$\begin{gathered} =59400-5 \\ =59395 \end{gathered}$$

Therefore, The smallest number which when increased by $\mathbf{5}$ is exactly divisible by the number $\mathbf{110}\mathbf{,}\mathbf{}\mathbf{135}$ and $\mathbf{200}$is $\mathbf{59395}$