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: <math> M \times \,^{\prime\prime} 0 \; 0 \; 1 \; 1 \; 1 \; 1 \; 1 \; 0 \,^{\prime\prime} = M \times (2^5 + 2^4 + 2^3 + 2^2 + 2^1) = M \times 62 </math>
where M is the multiplicand. The number of operations can be reduced to two by rewriting the same as
: <math> M \times \,^{\prime\prime} 0 \; 1 \; 0 \; 0 \; 0 \; 0 \; 0 \; 0 \mbox{-1} \; 0 \,^{\prime\prime} = M \times (2^6 - 2^1) = M \times 62. </math>
In fact, it can be shown that any sequence of 1s in a binary number can be broken into the difference of two binary numbers:
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