The smallest number which when increased by $17$ is exactly divisible by both $520$ and $468$ is:
(a) $4663$
(b) $4680$
(c) $4697$
(d) $4646$
Step-by-Step Solution
Key Concept: The required number is $\text{LCM}(520, 468) - 17$.
Prime factorisations: $520 = 2^3 \times 5 \times 13$, $468 = 2^2 \times 3^2 \times 13$. [0.5 Mark]
$\text{LCM}(520, 468) = 2^3 \times 3^2 \times 5 \times 13 = 8 \times 9 \times 5 \times 13 = 4680$.
Required number $= 4680 - 17 = 4663$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Finding $\text{LCM}(520, 468) = 4680$: 0.5 Mark
Subtracting 17 to get 4663: 0.5 Mark
Correct Answer: $4663$