Question

What is the smallest number which when divided by $15,20,36$ and $48$ gives the remainder $5$ in each case ?

Answer 1

Given , the numbers are $15,20,36$ and $48$

For finding the smallest number finding LCM of the given number .

So , Prime factors of the given number are

$15=3\times 5$

$20=2\times 2\times 5$

$36=2\times 2\times 3\times 3$

$48=2\times 2\times 2\times 2\times 3$

LCM of the numbers are $=2\times 2\times 2\times 2\times 3\times 3\times 5$

LCM of the numbers are $=720$

The smallest number gives remainder $5$ in each case is $720+5=725$

Hence, the required smallest number is $725$ .