IRL ABooklet

16.
Let $d = |A - B| = |B - C| = |C - D|$ then $d$ can take the values $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$. For $d = 0$ we find 9 numbers $1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999$. For other values of $d$, we take each of the eight sequences
$$+++, ++-, +−+, +−−, −+++, −+−, −−+, −−−$$
and find possible starting points to produce numbers with that particular sequence indicating for each digit whether the next digit is smaller or larger.

Case 1. `+++` (9 solutions)
$d = 1$ gives $1234, 2345, 3456, 4567, 5678, 6789$
$d = 2$ gives $1357, 2468, 3579$
$d = 3, 4, 5, 6, 7, 8, 9$ do not give further solutions.

Case 2. `++−` (16 solutions)
$d = 1$ gives $1232, 2343, 3454, 4565, 5676, 6787, 7898$
$d = 2$ gives $1353, 2464, 3575, 4686, 5797$
$d = 3$ gives $1474, 2585, 3696$
$d = 4$ gives $1595$

Case 3. `+-+` (36 solutions)

Case 4. `+--` (20 solutions)
Case 5. `-++` (20 solutions)
Case 6. `-+-` (45 solutions)
Case 7. $--+$ (20 solutions)
$d=1$ gives $2101, 3212, 4323, 5434, 6545, 7656, 8767, 9878$
$d=2$ gives $4202, 5313, 6424, 7535, 8646, 9757$
$d=3$ gives $6303, 7414, 8525, 9636$
$d=4$ gives $8404, 9515$
$d=5, 6, 7, 8, 9$ do not give further solutions.

Case 8. $---$ (12 solutions)
$d=1$ gives $3210, 4321, 5432, 6543, 7654, 8765, 9876$
$d=2$ gives $6420, 7531, 8642, 9753$
$d=3$ gives $9630$

Adding the 9 with $d=0$ to the count for each $+-$ pattern, we obtain
$9 + 9 + 16 + 36 + 20 + 20 + 45 + 20 + 12 = 187$.

Let $d = B - A$ so $d$ can take integer values from $-9$ to $9$ inclusive. If $d = 0$ then there are $9$ cases. So let us consider $d \neq 0$. We will first count the cases including those where $A = 0$ is permitted and then subtract those cases starting with zero.
When $d \neq 0$, there are four possible patterns for the three differences, namely
* Case $(d, -d, d)$. This requires two distinct digits separated by a gap $|d|$, with $10 - |d|$ cases.
* Cases $(d, d, -d)$ and $(d, -d, -d)$. This requires three distinct digits in arithmetic progression, with $10-2|d|$ cases if $d \le 4$ and none otherwise.
* Case $(d, d, d)$. This requires four digits in arithmetic progression with gap of $|d|$, i.e. the smallest digit $a$ must satisfy $a+3|d| \le 9$, which gives $10-3|d|$ cases if $|d| \le 3$ and no cases otherwise.
Now let us enumerate the cases with initial digit $A = 0$ and $d \neq 0$. There are no examples with $d < 0$. If $d > 0$, the difference pattern $(d, -d, -d)$ is impossible as it results in a negative final digit. The pattern $(d, -d, d)$ has solutions for any $1 \le d \le 9$, the pattern $(d, d, -d)$ has solutions for any $1 \le d \le 4$, the pattern $(d, -d, -d)$ has no solutions as the final digit cannot be negative, while the case $(d, d, d)$ has a solution if $1 \le d \le 3$.

We can now summarise all the cases according to the value of $|d|$ in the following table.

The solution to the problem then is 187 cases.

Let $d = |A - B| = |B - C| = |C - D|$.
For each value of $d$ from 0 to 9, set up a tableau with 4 rows (numbered 1 to 4) and 10 columns (numbered 0 to 9). The number $n(r, c)$ in row $r$ and column $c$ is the number of $r$-digit numbers, ending in the digit $c$, that have an absolute difference of $d$ between adjacent digits.

In row 1, for all $d$ we have $n(1,0) = 0$ because there is no 1-digit number ending in zero, while $n(1,c) = 1$ for $1 \le c \le 9$. In subsequent rows, by considering appending a final digit to a number of $r-1$ digits, $n(r,c)$ is the sum of $n(r-1,c-d)$ and $n(r-1,c+d)$ when $d > 0$. These summands are taken to be zero if the reference overspills the tableau, i.e. if $c - d < 0$ or $c - d > 9$. In the special case $d = 0$, the rows are all equal to the first row. The tableaux are shown below.

The total of the totals is 187.