Originally posted by googlefudge
You're missing some steps/instructions out.
What you have written makes no sense.
9(x+1) where x is an integer between 1 and 9 inclusive gives;
As the set of answers... Not 9.
Adding the first and last digit together from each number does indeed give
the answer 9 for each value of x... But nowhere do yo ...[text shortened]... it, about base 5 ... I still think you need to be clearer about
what your problem actually is.
Edited and inserted the final step which is to add the digits of the final result as you deduced that it should be.
As for the base 5 comments.
If you work in base 5 the available digits are: 0,1,2,3,4.
Hence the analogous expression to 9*(x+1) would be 4*(x+1).
Remember that in base 5 4+1=0 and you should carry 1 just like in base 10 (our familiar decimal system) it is 9+1=0 and you should carry 1.
So for instance pick x=2 in our base 5 system. The expression is:
4*(2+1) = 4*3
If you're not familiarized with multiplying integer numbers in other bases other than 10, the trick is just to see multiplication as a repeated addition.
Hence 4*3 is the same as 3+3+3+3. Remember our initial rule that 4+1=0 and carry 1 so that the result is
3+3= 1 carry 1
1+3=4 carry 0
4+3=2 carry 1
Hence the right most digit is 2 and the left most digit is the sum of the numbers you need to carry: 1+0+1=2 (the exact same method you use to sum integers in base 10!).
Thus the final result is 4*(2+1)=4*3=22 in base 4.
If you try it out in base 6 and write out a similar expression for the problem you'll see that the sum of final digits will always be 5 (6-1).
Did that made my comments about base 5 more intelligible?
PS: Even though the problem only states for one to chose numbers between 1 to 9 it is obvious that one can choose from 0 to 9. Not so obvious but also possible is for one to choose numbers greater than 9. The thing is that you have to sum the final digits more than one time...