$20$ -element subsets of $\{0,1,2,...,1024\}$ can be ordered to form a geometric progression modulo $1025$ with common ratio $2?$

How many
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Details and assumptions
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The sequence $\{a_i \}$ is a geometric progression modulo $1025$ with common ratio $2$ if $a_i \equiv a_1 \times 2^{i-1} \pmod{1025}$ .

The answer is 51.

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Note that $1025=5^2\cdot 41.$ Since $2^{10}\equiv -1 \pmod {1025},$ the order of $2$ in the multiplicative group modulo $41$ is $20.$ It is easy to check that the order of $2$ in the multiplicative group modulo $25$ is also $20$ . This implies that if $a,2a,4a,...,2^{19}a$ is a geometric progression modulo $1025,$ then $2^{20}a=a,$ and the progression can be "rotated" to start from every $2^i a.$

The numbers $1,2,...,2^{19}$ are distinct modulo $41$ and modulo $25,$ but they are not distinct modulo $5.$ Indeed, $2^4\equiv 2^0 \pmod 5.$ So if $a\equiv 0 \pmod {5\cdot 41},$ then $a\equiv 2^4a \pmod {1025},$ and we do not get a $20$ -term geometric progression. There are $5$ such residues $a$ modulo $1025.$ All other $a$ do give a geometric progression of $20$ distinct residues: if $a\neq 0 \pmod{41},$ then they are distinct modulo $41$ and if $a\equiv 0 \pmod{41},$ but $a\neq 0 \pmod{5},$ then they are distinct modulo $25.$ Note that, from the discussion in the first paragraph, each set gets counted $20$ times, so the answer is $\frac{1025-5}{20}=51.$