# 31st IMO 1990 shortlist p1

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'

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.

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