Fons wrote:But this one isn't working below 10, because the 0 should also be the symbol you get at the end.
Actually, it works fine also below 10. Here's the maths, first my earlier message:
I wrote:If
the first digit = a (1≤a≤9) and
the second digit = b (0≤b≤9),
then the 2-digit number =10a+b.
The calculation thus becomes
10a+b-(a+b) = 10a+b-a-b = 9a = {9,18,27,...,81}.
And then a slight modification:
If
the first digit = a (
0≤a≤9) and
the second digit = b (0≤b≤9),
then the 2-digit number =10a+b.
The calculation thus becomes
10a+b-(a+b) = 10a+b-a-b = 9a = {
0, 9,18,27,...,81}.
So, the calculation works, but the discrepancy lies in the wording of the task. It tells you to choose a 2-digit number and thus excludes 1-digit numbers. Furthermore, the symbol for 0 is different from that for 9, 18, 27, etc. but it
should be the same.
But the trick would of course be trivial if you were told to take a 1-digit number and subtract if from itself...
