Post by Stabbins McGee, simple country mathematician 🇵🇸 / Redsky

Stabbins McGee, simple country mathematician 🇵🇸 • Feed

𝘍𝘪𝘯𝘢𝘭𝘭𝘺 slaughtered this white whale. This problem was considered so difficult, that its difficulty made it to the news in Korea. You would sooner see a unicorn dig up Jimmy Hoffa than see a math problem on the news in the United States. Special thanks to 수학딜러 (www.youtube.com/watch?v=DF4d...).

YouTube explanation for this problem with the caption, 이 문제는 뉴스에서, 다른 유튜버들의 말처럼 욕할만큼의 킬러문제는 아니었어요 (평가원이 잘 만든 문제에요)

sep 2, 2025, 5:14 am • 2 0

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I miswrote it in the OP, but even the correct version is really difficult. Here's what I got, working off that video.

Suppose is a cubic function with a leading coefficient of that has the following property. For the function , there does not exist an integer such that . If and , find the value of . Consider the shape of the graph of From the description, we know we have a cubic function, and its local maximum must occur to the left of its local minimum. If we set , we can find that the derivative is always non-negative when . These are not the cubic functions we are looking for, as we know we need a negative derivative at . We only want cubic functions where . The condition has no effect on the end behavior of our cubic function. If is a large negative number, then is the product of two large negative numbers, which is positive. Similarly, if is a large positive number, then is the product of two large positive numbers, which is positive. We are concerned about the region near and between the local maximum and local minimum. There must exist a largest integer such that . In that case, the only way to meet the condition is to enture that . By the same token, there must exist a smallest positive number such that , so . In order to make the condition work, there must exist at least two zeros at adjacent integers. At least one zero must be . Identify the number of distinct zeros Consider cases of two distinct zeros If we had two distinct zeros, then or . In the former case, if , then and , so that's not what we want. In the latter case, creates a quadratic equation of that has no real solutions. Consider cases of three distinct zeros If there exist three distinct zeros, can be neither the leftmost nor the rightmost zero, as this would make it so that . We arrive at three cases. We'll list the roots in ascending order.

Consider cases of three distinct zeros If there exist three distinct zeros, can be neither the leftmost nor the rightmost zero, as this would make it so that . We arrive at three cases. We'll list the roots in ascending order. Case 1: zeros at In this case, . This one does not work because . Case 2: zeros at Here, . If , then when , , which violates our condition. Thus, if this works at all, then . Then only holds when , in which case . This case won't work. Case 3: zeros at Here, For reasons similar to the previous case, . Then only holds when , in which case , so we finally have a working case! This means , so . Source(s): 1 [+] (Korean).

sep 2, 2025, 5:14 am • 0 0 • view

I think this is for a different, but also difficult problem? I remember having thought and discussed about the screenshotted problem some time ago.

sep 2, 2025, 6:28 am • 1 0 • view

Oh dang, you’re right. I misread as I searched back!

sep 2, 2025, 7:00 am • 1 0 • view

I’ve really got to improve at how I organize these on BlueSky! I’ve been working on that on MyOpenMath, but I didn’t decide to adapt these systematically, problem by problem, until recently. I think I’ll use the MathSky tag more often, maybe!

sep 2, 2025, 7:03 am • 1 0 • view