## Task $5 / 83$

Let $n=\sum_{i=0}^{5} a_{i} \cdot 10^{i}$ be a 6-digit natural number, where $0 \leq a_{i} \leq 9, a_{i} \in N$, and $Q(n)=\sum_{i=0}^{5} a_{i}$ be its digit sum, and it holds that: $1. a_{i}>a_{k}$ for $i>k ; 2.10^{98}<n^{Q(n)}<10^{99}$.

Which numbers $n$ satisfy these conditions?
Due to the strict monotonicity of the logarithmic function in the interval $(0 ;+\infty)$, it follows from the 2nd condition $980$

$$
\frac{98}{Q(n)}<\lg n<\frac{99}{Q(n)}
$$

Due to the 1st condition, for $n$: $543210<n<987644 ; 5.7<\ln n<6.0$. From (1) and (2) it follows that

$$
\frac{98}{Q(n)}<6.0 \quad \text{and} \quad 5.7<\frac{99}{Q(n)}
$$

thus $16.3<Q(n)$ and $Q(n)<17.4$. Since $Q(n) \in \mathbb{N}$, it follows that $Q(n)=17$. From (1) it then follows that

$$
5.75<\frac{98}{17}<\lg n<\frac{99}{17}<5.83
$$

and thus $575439<n<676083$. In this interval, there are six numbers that satisfy condition 1: $654321 ; 654320 ; 654310 ; 654210 ; 653210 ; 643210$.

Of these, only $n=653210$ has the required cross sum of 17.