2. a) (3p) Determine the real values of $x$ for which

$\frac{\sqrt{2^{1980}-2^{1979}-\ldots-2^{1002}}+\sqrt{4+2^{2}+2^{3}+\ldots+2^{999}}}{\sqrt{x^{2}-1}}=\sqrt{32^{200}\left(x^{2}-1\right)}$.

Prof. Zanfirica Boiciuc

b) (4p) Determine the natural numbers written in base 10 with the property that each is 47 times larger than the sum of its digits.

Gazeta Matematică Nr.1/2014
## Solution:

a) $2^{1980}-2^{1979}-2^{1978}-\ldots-2^{1002}=2^{1980}-2^{1002}\left(2^{977}+2^{976}+\ldots+2^{2}+2+1\right)=2^{1980}-2^{1002} \cdot\left(2^{978}-1\right)=2^{1002}$

$4+2^{2}+2^{3}+\ldots+2^{999}=1+2^{0}+2^{1}+2^{2}+\ldots+2^{999}=1+2^{1000}-1=2^{1000}$

$2^{501}+2^{500}=32^{100} \sqrt{\left(x^{2}-1\right)^{2}} \Leftrightarrow 2^{500} \cdot 3=2^{500} \cdot\left|x^{2}-1\right| \Leftrightarrow\left|x^{2}-1\right|=3 \quad$ and using the condition $x^{2}-1>0 \Rightarrow x= \pm 2$.

b) Let $n$ be the number we are looking for and denote by $s(n)$ the sum of its digits. Clearly $n \geq 1 \cdot 47$, so $n$ has at least two digits. Let $\mathrm{n}=\overline{a_{1} a_{2} \ldots . a_{k}}, \mathrm{k} \geq 2$. We will have $10^{k-1} \leq n=47 \mathrm{~s}(n) \leq 47 \cdot 9 k=423 k$. The inequality $10^{k-1} \leq 423 k$ holds for $\mathrm{k}0 \Rightarrow x= \pm 2$ | $\mathbf{1} \mathbf{p}$ |
| b) Prove that the number sought can have 2, 3, or 4 digits | $\mathbf{1} \mathbf{p}$ |
| Prove that n is divisible by 9 | $\mathbf{1} \mathbf{p}$ |
| Prove that $\mathrm{n} \in \mathrm{M}_{423}$ | $\mathbf{1} \mathbf{p}$ |
| $\mathrm{n} \in\{423,846\}$ | $\mathbf{1} \mathbf{p}$ |