1. (AUS 3) The integer 9 can be written as a sum of two consecutive integers: $9=4+5$. Moreover, it can be written as a sum of (more than one) consecutive positive integers in exactly two ways: $9=4+5=2+3+4$. Is there an integer that can be written as a sum of 1990 consecutive integers and that can be written as a sum of (more than one) consecutive positive integers in exactly 1990 ways?
1. Let $N$ be a number that can be written as a sum of 1990 consecutive integers and as a sum of consecutive positive integers in exactly 1990 ways. The former requirement gives us $N=m+(m+1)+\cdots+(m+1989)=$ $995(2 m+1989)$ for some $m$. Thus $2 \nmid N, 5 \mid N$, and $199 \mid N$. The latter requirement tells us that there are exactly 1990 ways to express $N$ as $n+(n+1)+\cdots+(n+k)$, or equivalently, express $2 N$ as $(k+1)(2 n+k)$. Since $N$ is odd, it follows that one of the factors $k+1$ and $2 n+k$ is odd and the other is divisible by 2 , but not by 4 . Evidently $k+1<2 n+k$. On the other hand, every factorization $2 N=a b, 1<a<b$, corresponds to a The number of such factorizations is equal to $d(2 N) / 2-1$ because $a=b$ is impossible (here $d(x)$ denotes the number of positive divisors of an $x \in \mathbb{N})$. Hence we must have $d(2 N)=2 \cdot 1991=3982$. Now let $2 \mathrm{~N}=$ $2 \cdot 5^{c_{1}}-199^{c_{2}} \cdot p_{3}^{\epsilon_{1}} \cdots p_{\tau}^{\epsilon_{r}}$ be a factorization of $2 \mathrm{~N}$ into prime numbers, where $p_{3}, \ldots, p_{r}$ are distinct primes other than 2,5 , and 199 and $\epsilon_{1}, \cdots, \epsilon_{r}$ are positive integers. Then $d(2 N)=2\left(e_{1}+1\right)\left(e_{2}+1\right) \cdots\left(e_{r}+1\right)$, from which we deduce $\left(e_{1}+1\right)\left(e_{2}+1\right) \cdots\left(e_{r}+1\right)=1991=11 \cdot 181$. We thus get $\left\{e_{1}, e_{2}\right\}=\{10,180\}$ and $e_{3}=\cdots=e_{r}=0$. Hence $N=5^{10} \cdot 199^{180}$ and $N=5^{180} \cdot 199^{10}$ are the only possible solutions. These numbers indeed satisfy the desired properties.