Multiply by a constant with shifts

EasyProgramming & DSA~2m

A left shift n << k computes $2^k n$ . The constant $24$ can be written as a difference of two powers of two, so $24n = (n \ll a) - (n \ll b)$ with non-negative integer exponents $a > b$ (two shifts and one subtraction). Find $a + b$ .

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Write $24$ as a difference of two powers of two: the power just above $24$ is $32=2^5$ , and $32-24=8=2^3$ .
So $a=5$ , $b=3$ ; add them.

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Multiply by a constant with shifts

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