Choose two digits and arrange them to make two double-digit
numbers. Now add your double-digit numbers. Now add your single
digit numbers. Divide your double-digit answer by your single-digit
answer. Try lots of examples. What happens? Can you explain it?
Can you create a Latin Square from multiples of a six digit number?
When asked how old she was, the teacher replied: My age in years is
not prime but odd and when reversed and added to my age you have a
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
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:
For the one ending in 3:
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.