Find the least number $x$ such that $ 11$ divides $x$ and sum of its digits $S(x)$ is $27$.

Question

Find the least number $x$ such that $ 11$ divides $x$ and sum of its digits $S(x)$ is $27$.

Since $S(999)=27 $ it is clear that the number of digits $n>3.$ Let $x_i$ be digits then we have two equations \begin{cases} x_1+x_2+\cdots+x_n=27=5 \mod 11,\\ x_1+x_3+x_5+\cdots=x_2+x_4+\cdots \mod 11. \end{cases} It follows that $2(x_2+x_4+\cdots+)=5 \mod 11$ or $$ x_2+x_4+\cdots = 8 \mod 11. $$ For the case $n=5$ it reduced to $x_2+x_4=8$ and I manage to pick up the solution $10989.$

Is there better solution?

Answer 1

Since $\sum x_i=27$, $\sum (-1)^{i-1}x_i=27-2(x_2+x_4+\cdots)$ must be odd and divisible by $11$, and hence either $11$ or $-11$ (it must be between $-27$ and $27$.)

If $\sum (-1)^{i-1}x_i=11$ then $x_1+x_3+\cdots = 19, x_2+x_4+\cdots = 8$. The only way to get $19$ is with three digits, $(x_1,x_3,x_5)=(1,9,9)$ yielding the smallest values in lexicographical order. Then $x_2+x_4=8$ has $(0,8)$ for the smallest lexicographical order.

So you get $10989$.

If $\sum (-1)^{i-1}x_i=-11$, then $x_2+x_4+\cdots = 19$ and thus that there must be at least $6$ digits.

Answer 2

When $n = 5$, $10989$ is already the least solution to the problem.

First, when $n \le 4$, no solution exists. For $n = 3$, we deduce from the sum of digits $S(x) = 27$ that $x = 999$, which is not a multiple of $11$. For $n=4$, we have $x_2 + x_4 \equiv 8 \pmod{11}$.

Since each digit is an integer from $0$ to $9$ inclusive, we have $$0 \le x_n \le 9 \forall n \in \Bbb N^*,$$

from which we deduce that $x_2 + x_4 = 8$. Due to the sum of digits $S(x) = 27$, we have $x_1 + x_3 = 19 > 18$, which is clearly false.

We conclude that no solutions exist when $n \le 4$.

Now we come back to $n = 5$. We're given the number $10989$ which satisfies the given conditions, and we hope to prove that it's the minimal solution to the problem. From the condition in the OP $$x_2+x_4 (+\cdots) = 8 \pmod{11}.$$ and by observing the second and fourth digit, $0$ and $8$ are already the best combination. Since it's a five digit number, the leftmost digit is at least one. So we're left with two digits. By considering the sum of digits $S(x) = 27$, we know that the sum of the remaining two digits are $27 - 1 - 8 = 18$. This forces us to choose $9$ for the remaining two digits. Therefore, $10989$ is already the least natural number satisfying the given conditions in the OP.