Question

Find the smallest number which when increased by $15$ is exactly divisible by $448and520$

Answer 1

The smallest number which when increased by $15$ is exactly divisible by both $448and520$ is obtained by subtracting $15$from the LCM of $448and520.$

prime factorization of

$448=2\times 2\times 2\times 2\times 2\times 2\times 7$

$520=2\times 2\times 2\times 5\times 13$

$LCMof448and520=2^6\times 5\times 7\times 13=29120$

the required number

$$\begin{gathered} =(LCMof448and520)-15 \\ =29120-15 \\ =29105 \end{gathered}$$