Let's denote $S(x)$ as the sum of the digits of a natural number $x$. Solve the equations:

a) $x + S(x) + S(S(x)) = 1993$

b) $x + S(x) + S(S(x)) + S(S(S(x))) = 1993$.
a) According to the divisibility rule for 3, the numbers $x$ and $S(x)$ give the same remainder when divided by 3. The same remainder will also be given by the number $S(S(x))$. Therefore, the sum

$x + S(x) + S(S(x))$ is divisible by 3 (since it is the sum of three numbers with the same remainder when divided by 3). However, 1993 is not divisible by 3, so there are no solutions.

b) It is clear that $x < 1993$. It is not difficult to see that among the numbers less than 1993, the largest digit sum of 27 is achieved by the numbers 1989 and 999. Therefore, $S(x) \leq 27$. Further,

$S(S(x)) \leq S(19) = 10$. Finally, $S(S(S(x))) \leq 9$. From the equation, it follows that $x = 1993 - S(x) - S(S(x)) - S(S(S(x))) \geq$ $1993 - 27 - 10 - 9 = 1947$.

Similarly to a), all numbers $x, S(x), S(S(x)), S(S(S(x)))$ give the same remainder when divided by 9, and 1993 gives a remainder of 4, so $x \equiv 1 \pmod{9}$. Among the numbers from 1947 to 1993, the remainder 1 when divided by 9 is given by 1954, 1963, 1972, 1981, 1990. Checking these numbers, we find that only 1963 fits.

## Answer

a) No solutions; b) $x = 1963$.

Submit a comment