How many three digit numbers in base 11 are possible, whenexpressed in base 9, has it's digits reversed?

Question

*when expressed

Answer 1

This essentially becomes an algebra problem. If the three digits of a number in base 11 are x, y, and z, in that order, the number is equal to 121x + 11y + z. If the digits are reversed in base 9, the number is 81z + 9y + x. The two numbers are equal.

121x + 11y + z = 81z + 9y + x
120x + 2y = 80z

Since the digits occur in a base 9 number, each of them can only be 0, 1, 2, 3, 4, 5, 6, 7, or 8. I suggest you take each one for z and find the x and y (within the same set of 9 integers) that correspond to the choice, and discard any result where x or z is 0 (since you would end up with a 2 digit number for one of the combinations). Here are some examples.

Choose z = 0. This has no valid solutions because we reject z = 0.

Choose z = 1. 120x + 2y = 80*1 = 80. This has no valid solutions because x cannot be 0 and x = 1 gives 120 which is already greater than 80.

Choose z = 2. 120x + 2y = 80*2 = 160. This has no valid solutions because x = 1 gives 120 + 2y = 160 ==> y = 20, which is outside of the set of digits. x = 2 gives 240 which is greater than 160.

Choose z = 3. 120x + 2y = 80*3 = 240. This finally yields a valid solution with x = 2, y = 0. 203_base11 = 302_base9, which has a base 10 value of 245.

Answer 2

In addition to the previous answer which applies when z=3, there are solutions when z=6 and when z=9. When z=6:

406 base 11 is equal to 604 base 9 which is equal to 490 in base 10.

When z=9:

609 base 11 is equal to 906 base 9 which is equal to 735 in base 10.

The total answer is 3.