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Let Tag = x bits, Set = y bits, Block Offset = z bits.

x y z

So, x + y + z = 40.

Cache Size = 512KB = 219B.

Number of cache lines = $\frac{2^{19}}{2^{z}}$ = 2(19-z)

Given the cache is 8 way set associative.

So, number of sets = $\frac{2^{(19-z)}}{2^{3}}$ = 2(16-z)

From the first equation,

x + (16-z) + z = 40

So, x = 24 bits

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