31st IMO 1990 shortlist p1

Question

Question : Is there a positive integer which can be written as the sum of 1990 consecutive positive integers and which can be written as a sum of two or more consecutive positive integers in just 1990 ways? (IMO 1990 shortlisted)

I know that $N=m+(m+1)+...+(m+1989)=5×199×(2m+1989)$, but it's hard to me to solve ' written as the sum of 1990 consecutive positive integers'

Please help me.

Answer 1

If we sum up $a, a + 1, \dots, a + k$ with $a, k > 0$, then we get $(k + 1)a + \frac{k(k + 1)}2 = \frac 1 2(k + 1)(2a + k)$. Note that $k + 1 < 2a + k$ and they have different parities.

Conversely, given a number $N$ and a factorization $2N = uv$ with $1 < u < v$ and $u, v$ having different parities, we may solve out $k = u - 1$ and $a = \frac12(v - u + 1)$.

Therefore "the number of ways that $N$ can be written as a sum of two or more consecutive positive integers" is equal to "the number of ways that $2N$ can be written as a product $uv$ with $1 < u < v$ and $u\neq v \mod 2$".

Let's now consider the latter. We write $2N = 2^e t$ with $t$ odd. It is clear that, if we omit the condition $1 < u < v$, then the number of ways to write $2N$ as $uv$ with $u \neq v \mod 2$ is simply two times the number of divisors of $t$: every divisor $d$ of $t$ contributes to two ways, namely $(u, v) = (2^e d, \frac t d)$ or $(\frac t d, 2^e d)$.

Taking into account the condition $u < v$, we divide the result by $2$ and get just the number of divisors of $t$. Removing the case $u = 1$, we finally get the following:

If $N = 2^et$ with $m$ odd, then the number of ways that $N$ can be written as a sum of two or more consecutive positive integers is equal to $\tau(t) - 1$, where $\tau(t)$ stands for the number of divisors of $t$.

This being equal to $1990$, we see that $t$ has $1991$ divisors. Since $1991 = 11\times 181$, there are just two cases:

• $t = p^{10}q^{180}$ with $p, q$ primes.
• $t = p^{1990}$ with $p$ prime.

Now we also require that $N = 5 \times 199 \times (2m + 1989)$ for some integer $m$. This forces $N$ to be odd, and hence $e = 0$ and $N = t$.

Since both $5$ and $199$ divide $N$, the second case above is already impossible. Thus we are in the first case, and $p, q$ are $5$ and $199$, in some order.

This leads to two solutions:

• $5 \times 199 \times (2m + 1989) = 5^{10}199^{180}$, or
• $5 \times 199 \times (2m + 1989) = 199^{10}5^{180}$.

Both are obviously valid.

Hence the answer is "yes, there are in fact two such integers".