Digits, Bases and Periodicity

The digit sum is \(7+3+4+5+8+2+1=30\). Since \(30\equiv3\pmod9\), the original number also has remainder \(3\).

From the right, \(4-3+7-2+8-1+9=22\). Since \(22\) is divisible by \(11\), \(9182734\) is divisible by \(11\).

The last digits are \(3,9,7,1\), then the cycle repeats. Since \(2026\equiv2\pmod4\), the last digit is \(9\).

We have \((341)_b=3b^2+4b+1\).

Since \(1111=1+10+10^2+10^3\), we have \(1111=\frac{10^4-1}{9}\).

The digit sum is \(5+2+x+4+7=18+x\). This is divisible by \(9\) exactly for \(x=0\) or \(x=9\).

We get \(5-1+9-x+3-6=10-x\). We need \(10-x\equiv0\pmod{11}\). For a digit \(x\), this would require \(x=10\), impossible. Hence no such digit exists.

We have \(9^1\equiv9\), \(9^2\equiv81\), \(9^3\equiv29\), \(9^4\equiv61\), \(9^5\equiv49\), and \(9^{10}\equiv1\pmod{100}\). Since \(37\equiv7\pmod{10}\), \(9^{37}\equiv9^7\equiv69\pmod{100}\). The last two digits are \(69\).

We need \(3b+4\equiv0\pmod5\), so \(3b\equiv1\pmod5\). The inverse of \(3\) modulo \(5\) is \(2\), hence \(b\equiv2\pmod5\). With \(b>4\), \(b=5t+2\), \(t\ge1\).

We have \(10^2\equiv100\equiv9\pmod{13}\), \(10^3\equiv12\), \(10^4\equiv3\), \(10^5\equiv4\), \(10^6\equiv1\pmod{13}\). No smaller positive power gives \(1\), so the period is \(6\).

Since \(10^3=1000\equiv1\pmod{37}\), we have \(10^6\equiv1\pmod{37}\). Hence \(37\mid 10^6-1\). The number \(9\) is coprime to \(37\), so from \(R_6=\frac{10^6-1}{9}\) it follows that \(37\mid R_6\).

Each number is congruent to its digit sum modulo \(9\). The digit sums are equal, so the two numbers are congruent modulo \(9\). Therefore their difference is divisible by \(9\).

Because the number of digits is even, each pair of equal symmetric digits enters the alternating sum with opposite signs. Thus the alternating sum is \(0\), so the number is divisible by \(11\).

We have \(10^3=1000\equiv1\pmod{37}\), while \(10 ot\equiv1\) and \(10^2\equiv26 ot\equiv1\pmod{37}\). The order of \(10\) modulo \(37\) is \(3\). Hence \(37\mid R_n\) exactly when \(3\mid n\).

Since \(6\mid6n\), we have \(R_6\mid R_{6n}\). But \(R_6=111111=111\cdot1001\), and \(1001=7\cdot11\cdot13\). Hence \(R_{6n}\) is divisible by \(7\), \(11\), and \(13\).

Since \(3^{20}\equiv1\pmod{100}\) and \(7^{20}\equiv1\pmod{100}\), and \(2026\equiv6\pmod{20}\), we get \(3^{2026}\equiv3^6\equiv29\) and \(7^{2026}\equiv7^6\equiv49\pmod{100}\). Their sum is \(78\).

Checking residues \(b=0,1,\ldots,12\) gives \(b\equiv1\) or \(b\equiv3\pmod{13}\). With \(b>5\), the bases are \(b=13t+1\), \(t\ge1\), or \(b=13t+3\), \(t\ge1\).

We have \(10^1\equiv10\), \(10^2\equiv100\equiv19\), \(10^3\equiv190\equiv1\pmod{27}\). Smaller positive powers do not give \(1\), so the period is \(3\).

Since \(31\) is coprime to \(9\), the condition \(31\mid R_n\) is equivalent to \(10^n\equiv1\pmod{31}\). Computation gives \(10^5\equiv25\), \(10^{15}\equiv1\pmod{31}\), and the smaller divisors \(1,3,5\) of \(15\) do not give \(1\). Hence the order is \(15\), so \(31\mid R_n\) exactly when \(15\mid n\).

The numbers \(N\) and \(M\) have the same digit sum, hence they are congruent modulo \(9\). Therefore \(N-M\equiv0\pmod9\), so \(9\mid N-M\).

If one of \(R_1,\ldots,R_m\) is divisible by \(m\), we are done. Otherwise two residues coincide: \(R_i\equiv R_j\pmod m\), \(i

A number made of nines is always odd and ends in \(9\), so it is not divisible by \(2\) or \(5\). Thus the original statement is false. The corrected statement is: if \(\gcd(m,10)=1\), then such a number exists. Indeed, by the previous construction there is \(R_n\) divisible by \(m\); then \(9R_n\) consists only of nines and is also divisible by \(m\).

Since \(10^a\equiv1\pmod{R_a}\), for \(b=qa+r\), \(0\le r

We have \(2026=2\cdot1013\), and \(\gcd(1013,10)=1\). By the repunit construction, there exists \(R_n\) divisible by \(1013\). Then \(10R_n\) is obtained by appending a zero to a number of ones, so it consists only of digits \(0\) and \(1\). It is divisible by \(10\), hence by \(2\), and also by \(1013\). Therefore it is divisible by \(2026\).