Well done all those people who found the slippys ending in 2 and 3. How about slippy 4? That's a small one you can find quickly! Here is the best explanation of the process from Yeow Seng Poo, River Valley High School, Singapore:
I multiply the last digit of the slippy number, namely 2 or 3, by itself to get the last digit of the multiplied number that is also the second last digit in the slippy number. I then repeat the process with each of the digits, adding any tens produced during the multiplication of one digit to the next digit before continuing.
This continues till I reach an answer of 1 with no tens to bring over to the next possible digit, this is the answer. The last part is due to the fact that the first digit of the slippy number has to be one in order for the multiplied number to begin with 2 or 3.
Answers with workings:
For the one ending in 2:
|Slippy multiplied by 2:||2||1||0||5||2||6||3||1||5||7||8||9||4||7||3||6||8||4|
For the one ending in 3:
|Slippy number multiplied by 3:||3||1||0||3||4||4||8||2||7||5||8||6||2||0||6||8||9||6||5||5||1||7||2||4||1||3||7||9|
As for why there is only one sequence of digits in the slippy number ending in nine, this is because, going from units in increasing magnitude, each digit follows from the one before according to a fixed rule. However the cycle can be repeated to give slippys with twice as many digits or three times or more. The same is true for slippys ending in other digits.Try this for yourself with the slippys ending in 4.
The challenge is still out to programmers to send in programs to find slippy numbers.