Second Round

a. First we look at the last two digits of a sunny number. There are nine possibilities for these: $01$, $12$, $23$, $34$, $45$, $56$, $67$, $78$, and $89$. If we then look at twice a sunny number, we get the following nine possibilities, respectively, for the last two digits: $02$, $24$, $46$, $68$, $90$, $12$, $34$, $56$, and $78$. We see that twice a number can only be sunny if the original sunny number ends in $56$, $67$, $78$, or $89$. In all four cases we see that by doubling a $1$ carries over to the hundreds.

Now we look at the first two digits of a sunny number. The nine possibilities are $10$, $21$, $32$, $43$, $54$, $65$, $76$, $87$, and $98$. If the first digit is $5$ or higher, twice the number has more than four digits so it can never be sunny. The possibilities $10$, $21$, $32$, and $43$ are left. After doubling and adding the carried over $1$ to the hundreds we get, respectively, $21$, $43$, $65$, and $87$. In all cases twice a sunny number is a sunny number if the first digits of the original sunny number are $10$, $21$, $32$, or $43$ and the last two digits are $56$, $67$, $78$, or $89$. In total there are $4\cdot 4=16$ combinations to be made, hence $16$ sunny numbers for which twice the number is again sunny. $\square$

b. Denote by $a$ and $b$ the two middle digits of a sunny number. Then the two digits on the outside are $a+1$ and $b+1$, so the number is $1000(a+1)+100a+10b+(b+1)=1100a+11b+1001$. This number is divisible by $11$ because $100a$ as well as $11b$ as well as $1001=91\cdot 11$ is divisible by $11$. After division by $11$ we get the number $100a+b+91$. Now $b$ is at most $8$, because $b+1$ has to be a digit as well. Furthermore $a$ is at least $1$, because the number we started with has to be at least $2000$. So we see that $100a+b+91=100a+10\cdot 9+(b+1)$ is the three-digit number with digits $a$, $9$, and $b+1$, a three-digit number with a $9$ in the middle. $\square$