Library / /10 of 46

, 2015

Number theory Difficulty 5.0 AIME Prove it Japan

When five positive integers $a, b, c, d, e$ satisfy $a < b < c < d < e < a^2 < b^2 < c^2 < d^2 < e^2 < a^3 < b^3 < c^3 < d^3 < e^3$ , determine the minimum possible value that the sum $a+b+c+d+e$ can take.

Solution

From the given inequalities, it follows that $a+4 \le e$ and $e^2+1 \le a^3$ must hold. Therefore, we have
$(a+4)^2 \le e^2 \le a^3 - 1,$
from which it follows that $(a + 4)^2 \le a^3 - 1$ , i.e.,
$(a-4)(a^2+3a+4) \geq 1 > 0.$
This means that we must have $a > 4$ . Consequently, we have
$a+b+c+d+e > 5+6+7+8+9 = 35.$
On the other hand, we see that the choice of $(a, b, c, d, e) = (5, 6, 7, 8, 9)$ satisfies the requirements of the problem. Consequently, we conclude that $35$ is the desired answer.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.