Factors of $10^n-1$

Let $10^n-1\equiv 0\pmod{x}$ . This is equivalent to $10^n\equiv 1\pmod{x}$ .

However, note that by Euler's Theorem, $10^{\phi(x)}\equiv 1\pmod{x}$ as long as $\text{gcd}(x,10)=1$ . Since $\phi(x)$ clearly exists for all $x$ , we see that there exists an $n$ as long as $\text{gcd}(x,10)=1$ .

Now we must prove that when $\text{gcd}(x,10) \ne 1$ , then there does not exist an $n$ . The statement $\text{gcd}(x,10)\ne 1$ is equivalent to $n\equiv 0\pmod{2\text{ or }5}$ .

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Claim: when $x\equiv 0\pmod{2\text{ or }5}$ , then there does not exist an $n$ .

Proof: If $x\equiv 0\pmod{2}$ , then $x$ is even. But $10^n-1$ is odd, so there cannot exist an $n$ .

If $x\equiv 0\pmod{5}$ , then $10^n-1\equiv 0\pmod{5}$ . But $10^n-1\equiv 0^n-1\equiv 4\pmod{5}$ , so there does not exist any $n$ .

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We have proved that for all $x$ not a multiple of $2$ and $5$ , there exists an $n$ . Thus the probability is $\dfrac{4}{10}=\dfrac{2}{5}$ and our answer is $2+5=\boxed{7}$ .

The probability that a given positive integer x is a factor TO 10^n - 1 for some positive integer n is equal to A/ B where A and B are positive co prime integers. Find A + B. {Modified 'of' into 'TO' to eliminate ambiguity.}

10^n – 1: 9, 99, 999, 9999….

10^n – 3: 7, 97, 997, 9997…

10^n – 7: 3, 93, 993, 9993…

10^n – 9: 1, 91, 991, 9991…

10^n – 11: 88, 988, 9988… …

Same answer of 0.4 or 2/ 5 for 7; this is true for either one of the above, provided n can keep on to infinity.

9999…9999 and 9999…9991 are the same in infinity. Excluding MOD 2 and MOD 5 means excluding MOD 10 and the remained are all inclusive in 10^n – 1.

1.000000000000000000 1 1, 9.00000000000000000E+0000 ONE!

0.500000000000000000 1 2, 0.00000000000000000E+0000

0.666666666666666667 2 3, 9.00000000000000000E+0000 ONE!

0.500000000000000000 2 4, 0.00000000000000000E+0000

0.400000000000000000 2 5, 0.00000000000000000E+0000

0.333333333333333333 2 6, 0.00000000000000000E+0000

0.428571428571428571 3 7, 9.99999000000000000E+0005 ONE!

0.375000000000000000 3 8, 0.00000000000000000E+0000

0.444444444444444444 4 9, 9.00000000000000000E+0000 ONE!

0.400000000000000000 4 10, 0.00000000000000000E+0000

0.454545454545454545 5 11, 9.90000000000000000E+0001 ONE!

0.416666666666666667 5 12, 0.00000000000000000E+0000

0.461538461538461538 6 13, 9.99999000000000000E+0005 ONE!

0.428571428571428571 6 14, 0.00000000000000000E+0000

0.400000000000000000 6 15, 0.00000000000000000E+0000

0.375000000000000000 6 16, 0.00000000000000000E+0000

0.411764705882352941 7 17, 9.99999999999999900E+0015 ONE!

0.388888888888888889 7 18, 0.00000000000000000E+0000

0.421052631578947368 8 19, 9.99999999999999999E+0017 ONE! ADDED IN.

0.450000000000000000 9 20, 0.00000000000000000E+0000

0.476190476190476190 10 21, 9.99999000000000000E+0005 ONE!

0.454545454545454545 10 22, 0.00000000000000000E+0000

0.434782608695652174 10 23, 9.99999999999999999E+0017 ? EXCEED! ADDED IN.

0.458333333333333333 11 24, 0.00000000000000000E+0000

0.440000000000000000 11 25, 0.00000000000000000E+0000

…

Eventually 0.4 if limited significant figures of computer not exceeded. To overcome machine limitation, modify the routine, simplified task tells an exact 0.4 at every check of x MOD 100000 = 0 for example. To explain in short, we are actually only finding the probability of integer x which are not multiple of 2 and not multiple of 5 given arbitrarily.

A + B = 2 + 5 = 7.