Question

Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

• A. 4663
• B. 2244
• C. 7893
• D. 4416

Answer 1

The correct option is A 4663

Let the required number is $x.$
$\Rightarrow (x+17)$ is the smallest number divisible by 520 and 468.
$\Rightarrow (x+17)=\text{LCM}(520,468)$

Prime factorisation of 520 and 468:
$520=2^3\times 5\times 13$
$468=2^2\times 3^2\times 13$

$\Rightarrow \text{LCM}=2^3\times 3^2\times 5\times 13=4680$

Thus, $x+17=4680.$
$\Rightarrow x=4663$

Therefore, the required number is 4663.