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do we have anything better than this

What do u mean by better?

If u shift the bits 3 times, u r multiplying the number by 2^3 = 8. so now subtract the number by itself to multiply it by 7 times.

perfect

just perfect... simple logic.... efficient use...!!!!

Respect!

Say you have to multiply 'a' by 'b'.
Find a smallest number 'c' which is greater than 'b' and is a power of 2 (you would be interested in this because you would want to shift bits - efficient method.) so if c = 2^k then a.c = (a << k) right shifting 'a' by k bits.

Now, we know that a.c is greater than a.b, so we would want to subtract the difference [(c-b).n] off the *product*,

simply put: b = ( c - delta ) -> a.b = (c-delta) . a -> a.b = a.c - a.delta

[delta is known in all cases since you pick 'c' and 'b' is already given]

explanation for the first soln....
logic :
7*x = 8*x - x
7*x = (x<<3) - x

as x<<3 = x*(2^3);
case closed ;)