XIII OM - II - Task 6

Find a three-digit number with the property that the number represented by these digits and in the same order, but in a different base of numeration than $ 10 $, is twice as large as the given number.
If $ x $, $ y $, $ z $ denote the consecutive digits of the sought number, and $ c $ - the new base of numeration, then

the equation can be replaced by the equation

The task is to solve equation (1) in integers $ x $, $ y $, $ z $, $ c $, satisfying the conditions $ 1 \leq x \leq 9 $, $ 0 \leq y \leq 9 $, $ 0 \leq z \leq 9 $, $ c > 0 $.
We will first prove that the only possible value of $ c $ is $ 15 $. Indeed, when $ x $, $ y $, $ z $ satisfy the above conditions, and $ 0 < c \leq 14 $, then

while if $ c > 16 $, then

Substituting $ c = 15 $ into equation (1), we obtain the equation

If integers $ x $, $ y $, $ z $ satisfy equation (2), then $ z $ is divisible by the common factor $ 5 $ of the first two terms, and since $ 0 \leq z \leq 9 $, there are only two possibilities:
a) $ z = 0 $; from equation (2) we get

from which it is clear that $ x $ and $ y $ can only have the values $ x = 1 $, $ y = 5 $, which gives the sought number the value $ 150 $. This number indeed satisfies the condition of the problem, as in base $ 15 $ numeration it represents the number $ 15^2 + 5 \cdot 15 = 300 = 2 \cdot 150 $.
b) $ z = 5 $; to determine $ x $ and $ y $ we get from (2) the equation:

from which it is clear that $ x < 3 $, since for $ x \geq 3 $ we have $ -5x + y + 1 \leq -15 + 9 + 1 = -5 $. When $ x = 1 $, then $ y = 4 $, and when $ x = 2 $, we have $ y = 9 $; the corresponding values of the sought number are $ 145 $ and $ 295 $. Both these numbers satisfy the condition of the problem, as

The problem thus has $ 3 $ solutions, which are the numbers $ 145 $, $ 150 $, $ 295 $. The appropriate base of numeration for each of them is $ 15 $.