91Ó°ÊÓ

Obtain the two incongruent solutions modulo 210 of the system $$ \begin{aligned} &2 x \equiv 3(\bmod 5) \\ &4 x \equiv 2(\bmod 6) \\ &3 x=2(\bmod 7) \end{aligned} $$

(a) Given an integer \(N\), let \(M\) be the integer formed by reversing the order of the digits of \(N\) (for example, if \(N=6923\), then \(M=3296\) ). Verify that \(N-M\) is divisible by \(9 .\) (b) A palindrome is a number that reads the same backwards as forwards (for instance, 373 and 521125 are palindromes). Prove that any palindrome with an even number of digits is divisible by 11 .

If \(a \equiv b\left(\bmod n_{1}\right)\) and \(a \equiv c\left(\bmod n_{2}\right)\), prove that \(b \equiv c(\bmod n)\), where the integer \(n=\operatorname{gcd}\left(n_{1}, n_{2}\right)\)

Obtain the eight incongruent solutions of the linear congruence \(3 x+4 y \equiv 5(\bmod 8)\).

Given a repunit \(R_{n,}\), show that (a) \(9 \mid R_{n}\) if and only if \(9 \mid n\). (b) \(11 \mid R_{n}\) if and only if \(n\) is even.

Find the solutions of each of the following systems of congruences: (a) \(5 x+3 y=1(\bmod 7)\) \(3 x+2 y=4(\bmod 7)\) (b) \(7 x+3 y=6(\bmod 11)\) \(4 x+2 y \equiv 9(\bmod 11)\) (c) \(11 x+5 y=7(\bmod 20)\) \(6 x+3 y \equiv 8(\bmod 20)\)

Factor the repunit \(R_{6}=111111\) into a product of primes. [ Hint: Problem \(16(a) .]\)

Explain why the following curious calculations hold: $$ \begin{aligned} 1 \cdot 9+2 &=11 \\ 12 \cdot 9+3 &=111 \\ 123 \cdot 9+4 &=1111 \\ 1234 \cdot 9+5 &=11111 \\ 12345 \cdot 9+6 &=111111 \\ 123456 \cdot 9+7 &=1111111 \\ 1234567 \cdot 9+8 &=11111111 \\ 12345678 \cdot 9+9 &=111111111 \\ 123456789 \cdot 9+10 &=1111111111 \end{aligned} $$ [Hint: Show that $$ \begin{aligned} &\left(10^{n-1}+2 \cdot 10^{n-2}+3 \cdot 10^{n-3}+\cdots+n\right)(10-1) \\ &\left.+(n+1)=\frac{10^{n+1}-1}{9} \cdot\right] \end{aligned} $$

For any prime \(p>3\) prove that 13 divides \(10^{2 p}-10^{p}+1\). [Hint: By Problem \(\left.16(a), 10^{6}=1(\bmod 13) .\right]\)