Or search by topic
This solution to part (a) comes from Pierce Geoghegan:
$aabb = a(10^3)+ a(10^2) +b(10) + b = 11(b + 100a).$
Let $aabb = x^2.$ We know $x^2=0\ \rm{(mod\ 11)}$ and that implies $x^2 = 0\ \rm {( mod\ 121)}$ (since $11$ is prime $11^2$ must be a factor of $x^2$ ). So $x^2 = 121y^2$ for some integer $y$ and since $aabb < 10000$ we know $y< 10.$
Testing reveals $y=8$ and $aabb=7744$.
These solutions for part (b) use the Remainder Theorem and Modular Arithmetic. Can you prove the result using the Binomial Theorem, or yet another method?
This solution comes from Pierce Geoghegan.
For the second question note:
$11 = 11\ \rm{(mod\ 25)}$ ; $11^2 = -4\ \rm{(mod\ 25)}$ ; $11^3 = -19\ \rm{(mod\ 25)}$ ; $11^4 = 16\ \rm{(mod\ 25)}$ ; $11^5 = 1 \ \rm {( mod\ 25)}$ ... (A)
Now $11^{10}-1=(11^5 + 1)(11^5 - 1)$ and using (A)
$(11^5)- 1 = 0\ \rm {(mod\ 25)}$
Also $11^5 - 1=0\ \rm {(mod\ 2)}$ since $11^5=1\ \rm {(mod\ 2)}$ and $11^5 + 1=0\ \rm {(mod\ 2)}$ since $11^5=1\ \rm {(mod\ 2)}.$
So $(11^5 + 1)(11^5 - 1) = (0\ \rm{(mod\ 2)})(0\ \rm {(mod\ 50)}) = 0\ \rm {(mod\ 100)}.$ QED
This is Ang Zhi Ping's solution.
Let a polynomial $P(x)$ be $x^{10}-1$ . Observe that when $x=1$ , $P(x)$ is reduced to 0, ($1^{10}-1=0$). Hence, $(x-1)$ is a factor of $P(x)$ .
Thus, $11^{10}-1$ can be factorized to $(11-1)Q(x)$ , whereby $Q(x)$ is the quotient. We know by now, $11^{10}-1$ is a multiple of 10. Taking $x^{10}-1=(x-1)Q(x)$ , the quotient $Q(x)$ can be easily evaluated using synthetic division:
$x^{10}-1=(x-1)(x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)$
Hence
$Q(x)= (x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1).$
One can see that when $x=1$ , the remainder of $Q(1)=10$ . Thus, $Q(x)=(x-1)R(x)+10$ and $(R(x)$ is another quotient. For $11^{10}-1$ , $P(x)=(x-1)((x-1)R(x)+10)$ which gives $P(11)=(11-1)((11-1)R(x)+10)=10(10R(x)+10)=100(R(x)+1).$
Thus, $11^{10}-1=100(R(x)+1)$ , and hence the number is a multiple of $100$.