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The greatest 4-digit number such that when divided by $$16$$, $$24$$ and $$36$$ leaves $$4$$ as remainder in each case is
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The greatest 4-digit number such that when divided by $$16$$, $$24$$ and $$36$$ leaves $$4$$ as remainder in each case is
Let the number be $$n$$
The number reduced by $$4$$ must be a common multiple of $$16, 24, 36$$, so it is a multiple of their LCM.
So, $$(n-4)=k\times L.C.M[16,24,36]$$ ,where $$k$$ is some natural number
or, $$n=k\times L.C.M[16,24,36]+4$$
Since $$16 = 2^4$$, $$24 = 2^3 \times 3$$ and $$36 = 2^2 \times 3^2$$, the LCM is $$2^4 \times 3^2 = 144$$.
$$\therefore n=144k+4$$
Now, since $$n$$ is the greatest four-digit number that follows this pattern
So, we should take $$k=69$$ as any other value greater than $$69$$ makes $$n$$ a five digit number
$$\therefore n=144*69+4=9940$$
Hence, Option B is correct
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