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Find the greatest number less than $10,000$ which is exactly divisible by $48, 60$ and $65$.
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$48 = 2^4 \times 3$, $60 = 2^2 \times 3 \times 5$, $65 = 5 \times 13$ (1/2 Mark)
L.C.M. $(48,60,65) = 2^4 \times 3 \times 5 \times 13 = 3120$ (1 Mark)
$\therefore$ Highest multiple of $3120$, less than $10,000 = 3120 \times 3 = 9360$ (1/2 Mark)
L.C.M. $(48,60,65) = 2^4 \times 3 \times 5 \times 13 = 3120$ (1 Mark)
$\therefore$ Highest multiple of $3120$, less than $10,000 = 3120 \times 3 = 9360$ (1/2 Mark)