Question

Find the smallest number which when divided by $12,20,30$ or $60$ leaves a remainder $5$ each time.

Answer 1

We first need to find the LCM of $12,20,30$ and $60$.
To find LCM, we first write all numbers as products of their prime numbers.

$12=2\times 2\times 3$

$=2^2\times 3^1$

$20=2\times 2\times 5$

$=2^2\times 5^1$

$30=2\times 3\times 5$

$=2^1\times 3^1\times 5^1$

$60=2\times 2\times 3\times 5$

$=2^2\times 3^1\times 5^1$

We then choose each prime number with the greatest power and multiply them to get the LCM.

$\begin{array}{cc}LCM& =2\times 2\times 3\times 5\\ & =4\times 3\times 5\\ & =60\end{array}$

Hence, $60$ is the smallest number which is exactly divisible by $12,20,30$ or $60$

So, the smallest number which when divided by $12,20,30$ or $60$ leaves a remainder $5$ each time will be $60+5=65$.