An Eilitnip number is a three-digit positive integer with the properties that, for some integer $k$ with $0 \leq k \leq 7$ :

- its digits are $k, k+1$ and $k+2$ in some order, and
- it is divisible by $k+3$.

Determine the number of Eilitnip numbers.
We make a table that lists, for each possible value of $k$, the digits, the possible three-digit integers made by these digits, and $k+3$ :

| $k$ | $k, k+1, k+2$ | Possible integers | $k+3$ |
| :---: | :---: | :---: | :---: |
| 0 | $0,1,2$ | $102,120,201,210$ | 3 |
| 1 | $1,2,3$ | $123,132,213,231,312,321$ | 4 |
| 2 | $2,3,4$ | $234,243,324,342,423,432$ | 5 |
| 3 | $3,4,5$ | $345,354,435,453,534,543$ | 6 |
| 4 | $4,5,6$ | $456,465,546,564,645,654$ | 7 |
| 5 | $5,6,7$ | $567,576,657,675,756,765$ | 8 |
| 6 | $6,7,8$ | $678,687,768,786,867,876$ | 9 |
| 7 | $7,8,9$ | $789,798,879,897,978,987$ | 10 |

When $k=0$, the sum of the digits of each three-digit integer is 3 , so each is divisible by 3 .

When $k=1$, only two of the three-digit integers are even: 132 and 312. Each is divisible by 4 .

When $k=2$, none of the three-digit integers end in 0 or 5 so none is divisible by 5 .

When $k=3$, only two of the three-digit integers are even: 354 and 534 . Each is divisible by 6 . When $k=4$, the integer 546 is divisible by 7 . The rest are not. (One way to check this is by dividing each by 7 .)

When $k=5$, only two of the three-digit integers are even: 576 and 756 . Only 576 is divisible by 8 .

When $k=6$, the sum of the digits of each of the three-digit integers is 21 , which is not divisible by 9 , so none of the integers is divisible by 9 .

When $k=7$, none of the three-digit integers end in 0 so none is divisible by 10 .

In total, there are $4+2+2+1+1=10$ three-digit integers that satisfy the required conditions.

ANSWER: 10