New Year's Countdown Day 22: Will Newton Help?

Algebra Level 3

$\begin{cases} \displaystyle \sum_{i = 1}^{22} a_i^{22} = 22 \\ \displaystyle \sum_{i = 1}^{22} a_i^{23} = 20 \\ \displaystyle \sum_{i = 1}^{22} a_i^{24} = 17 \end{cases}$

Does there exist positive real numbers $a_1, a_2, a_3, \dots, a_{22}$ such that the above equations are satisfied?

This problem is part of the set New Year's Countdown 2017 .

Not enough information No Yes

1 solution

Chan Tin Ping
Dec 23, 2017

I use the theorem $f(n)=(\dfrac{\sum_{i=1}^{m}a_i^{n}}{m})^{\frac{1}{n}}$ is increasing function.

For the first case, which is $f(22)$ , we can get that $f(22)=(\dfrac{22}{22})^{\frac{1}{22}}=1$ . However, if we evaluate second case, we will get $f(23)=(\dfrac{20}{22})^{\frac{1}{23}}$ , which is obviously smaller than $1=f(22)$ . By the theorem, this is a contradiction. Hence, there dosen't exist such numbers.

New Year's Countdown Day 22: Will Newton Help?

This problem is part of the set New Year's Countdown 2017 .

1 solution

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