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Suppose I have two binary numbers which has a combination of both zeroes and ones. Is it possible to show that such two binary numbers which have at least one zero in it can never be multiplied to give a binary number with no zero in it ( i.e. , the resulting binary number is a combination of only 1's ) ?

Spectre
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Abhinav Tahlani
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  • Binary numbers that have only $1$s are $2^n-1$. $2^m-1$ divides $2^n-1$ when $m$ divides $n$, so some (but not all - see my answer) factors of $2^n-1$ have only $1$s – J. W. Tanner Feb 12 '20 at 05:58

2 Answers2

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No. Try $23\times89=2047$ in binary: $10111_2\times1011001_2=11111111111_2$.

J. W. Tanner
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Minimal counterexample: the binary number $11111111$.

Indeed, one has $(110011)_2 \times (101)_2 = (11111111)_2 = (51)_{10} \times (5)_{10} = (255)_{10}$ and the smaller binary numbers containing only $1$'s cannot be decomposed as a product of binary numbers containing a $0$: $(11)_2 = (3)_{10}, (111)_2 = (7)_{10}, (11111)_2 = (31)_{10}, (1111111)_2 = (127)_{10}$ are prime, $(1111)_2 = (15)_{10} = (11)_2 \times (101)_2$ and $(111111)_2 = (63)_{10} = (111)_2 \times (1001)_2$.

J.-E. Pin
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