C2. We call a positive integer sunny if it consists of four digits and if, moreover, each of the two outer digits is exactly 1 greater than the digit next to it. The numbers 8723 and 1001 are, for example, sunny, but 1234 and 87245 are not.

(a) How many sunny numbers are there whose double is also a sunny number?

(b) Prove that every sunny number greater than 2000 is divisible by a three-digit number that has a 9 in the middle.
C2. (a) We first look at the last two digits of a sunny number. For this, there are the following nine possibilities: $01, 12, 23, 34, 45, 56, 67, 78$ and 89. If we then look at the double of a sunny number, we get the following nine possibilities for the last two digits: $02, 24, 46, 68, 90, 12, 34, 56$ and 78. We see that the double can only be a sunny number if the original sunny number ends in 56, 67, 78, or 89. In all four cases, we see that there is also a 1 carried over to the hundreds place in the doubling.

$\mathrm{Now}$ we look at the first two digits of the sunny number. The nine possibilities are 10, 21, 32, 43, 54, 65, 76, 87, and 98. If the first digit is 5 or higher, then after doubling we have a total of five digits and we definitely do not get a sunny number. The remaining possibilities are 10, 21, 32, and 43. After doubling and adding the 1 to the hundreds place, we get 21, 43, 65, and 87, respectively. In all cases, the double of a sunny number is again a sunny number, as long as 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 we can make, and thus 16 sunny numbers for which the double is again a sunny number.

(b) Let the middle two digits of a sunny number be $a$ and $b$. Then the outer two digits 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 both $1100a$ and $11b$ and $1001=91 \cdot 11$ are divisible by 11. After division by 11, we get the number $100a+b+91$. Now, $b$ is at most 8, because $b+1$ must also be a digit. Furthermore, $a$ is at least 1, because the original sunny number must be greater than 2000. Therefore, we see that $100a+b+91=100a+10 \cdot 9+(b+1)$ is the three-digit number with the digits $a, 9$, and $b+1$, a number of three digits with a 9 in the middle.