Find maximal positive integer $p$ such that $5^7$ is sum of $p$ consecutive positive integers
Problem
Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2013
Tags: number theory, Maximal
Tintarn
11.10.2018 18:18
So $5^7=n+(n+1)+\dotsc+(n+p-1)=np+\frac{p(p-1)}{2}$ and hence $2 \cdot 5^7=p(2n+p-1)$ for some positive integer $n$. But this is equivalent to $p \mid 2 \cdot 5^7$ and $p(p-1)<2 \cdot 5^7$ Clearly $p \ge 5^4$ implies $p(p-1) \ge 5^4 \cdot (3 \cdot 5^3)=3 \cdot 5^7$ which is too large. Hence $p \le 2 \cdot 5^3$. On the other hand, $p=2 \cdot 5^3$ clearly works since then $p(p-1)<p^2=4 \cdot 5^6<5^7<2 \cdot 5^7$.
jeff10
25.02.2021 19:08
We first need $\frac{p(p+1)}{2} \le 5^7$, or $p(p+1) \le 156,250$. It follows that $(p+0.5)^2 \le 156,250.25 < 160,000 = 400^2$. Hence, $p < 400$.
We know that $p$ divides $5^7$ or $v_2{(p)} = v_2{(5^7)} + 1$. Hence, the only other possibility is that $p$ is an even number such that $p$ divides $2 \times 5^7$. The largest valid odd positive integer less than $400$ is $125$. The largest valid even positive integer less than $400$ is $250$. One can check that in both cases, $(p+0.5)^2$ is not even close to $156,250.25$. Hence, the desired answer is $\boxed{250}$.