NT with repeating decimal digits

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NT with repeating decimal digits

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Source: Romania EGMO TST 2019 Day 1 P2

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oVlad

#1 h Apr 21, 2025, 1:46 PM

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Determine the digits $0\leqslant c\leqslant 9$ such that for any positive integer $k{}$ there exists a positive integer $n$ such that the last $k{}$ digits of $n^9$ are equal to $c{}.$

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kokcio

#2 h Apr 21, 2025, 3:32 PM

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We cannot have $c=2,4,6,8$ , because last four digits would have to be divisible by $16$ , but in each of this cases numbers $2222, 4444, 6666, 8888$ are not. Similarly, we can see that $c=5$ leads to contradiction.
Now, assume that $c$ is coprime to $10$ . We know that $9$ is coprime with $\phi(5^n)$ and $\phi(2^n)$ for all $n$ , so function $f(x)=x^9$ is bijective on integers relatively prime to $10^n$ (we count modulo $10^n$ ). To see that this function is bijective, we can also see that $a^9\equiv b^9\mod 5^n$ iff $5^n$ divides $(a-b)(a^2+ab+b^2)(a^6+a^3b^3+b^6)$ , but $5$ cannot divide neither $a^2+ab+b^2$ , nor $a^6+a^3b^3+b^6$ if $ab$ is not divisible by $5$ , so we would have to have $a\equiv b\mod5^n$ . The same argument works modulo $2^n$ . Therefore, we can have $c=1,3,7,9$ . Obviously, we can also have $c=0$ , so our answer is $c\in\{0,1,3,7,9\}$ .

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