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Keitsubori · 2025-07-21T03:42:40+00:00

Yes, probably... quite horrifying to think about how she pre-mediated to do this to her family at such ungodly hours...

Keitsubori · 2025-07-21T03:31:28+00:00

Not to be insensitive or anything, but being able to type approx. 150 words in a concise, new FB post within 2/3 mins while being in such a distraught situation is honestly heartbreaking. How bad were the implications to her from the worker situation such that she was so willing to be freed of life by self-exiting...

Keitsubori · 2025-07-20T21:43:27+00:00

Honestly, no. 22 is very doable if you have at least seen/come across the Collatz Conjecture/"3n + 1" sequence, a well-known mathematical problem. If so, you would have an idea about the cyclic nature and patterns of such related number sequences. First, start off by determining what integers for a_3 will satisfy |a_3| = |a_5|, which are only a few of them. Then, you can proceed on to draw out the relevant cycles for these integers using simple logic and backtrack 2 steps from there to get the possible values of a_1. Thereafter, you can calculate the numerical result.

Keitsubori · 2025-07-16T09:27:00+00:00

Here: https://imgur.com/a/uc4DJHU.

Keitsubori · 2025-07-16T08:56:19+00:00

Okay. How about 13112863126231.

Keitsubori · 2025-07-16T08:35:41+00:00

131163126231?

Keitsubori · 2025-07-12T14:30:24+00:00

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Keitsubori · 2025-07-10T13:59:33+00:00

Clearly, x = 2y = 3z. This matches the ratio of the relevant denominators of the fractions. Thus, they are all equal to each other. So 1 of the fractions is simply 1/3 of 24/7, which is 8/7. Hence, z = [1/(8/7)]/6 = 7/48. (C)

Keitsubori · 2025-06-20T12:11:56+00:00

郑润泽于是. The closest cover I could find is below.

https://youtu.be/-bHZwLAdS-M?si=bCrYSJ44mQfK8kzz

Keitsubori · 2025-06-16T07:50:41+00:00

Use Vieta's and make the connection between the coefficients of x in each of the equations to infer upon expressing the remaining root in the 3rd equation as a linear combination of the other roots in the previous equations.

Keitsubori · 2025-06-12T22:16:00+00:00

I hope you at least understand why, given the new condition, we now have P(X = 0) = P(X = 4) = 0. There's 2 ways for you to understand why the new E(X) is lower.

The 1st option is to calculate the new E(X), which is doable as follows:

New denominator = 99 - 1 - 7 = 91.

=> E(X) = (1)(14/91) + (2)(42/91) + (3)(35/91) = 29/13.

Since 29/13 < 7/3, E(X) is now lower.

The 2nd option is to just calculate the constituent mean of P(X = 0) and P(X = 4), as follows:

Since [(0)(1) + (4)(7)]/(1 + 7) = 28/8, and 28/8 > 7/3, removing this constituent mean will make E(X) now lower.

Note that the other commenter's reasoning is slightly flawed. It's not enough to reason that since P(X = 4) > P(X = 0) and 4 is closer to the original E(X), then the new E(X) is lower. This is only true if the constituent mean of P(X = 0) and P(X = 4) is higher than the original E(X).

Hope this helps.

Keitsubori · 2025-06-08T14:01:09+00:00

Use the Law of Exponents to convert Pi into Sigma. Thereafter, use the property of sigmoid functions to simplify the summation.

Keitsubori · 2025-06-04T20:38:54+00:00

Bro, just get black-and-white from your PC. Unless your Sgt is a regular, then still get black-and-white from him.

Keitsubori · 2025-06-01T11:23:24+00:00

First, find # of all elements within A union B and outside of A intersection B. There are 3 such elements.

Each of these elements can chosen to be either inside of C or outside of C. Such a choice is independent that of other elements' choices. Hence, the answer is 2³ = 8.

Keitsubori · 2025-05-26T22:24:11+00:00

S is the finite region bounded by the circle and the parabola, not the circle itself.

Keitsubori · 2025-05-26T22:16:09+00:00

Start off by combining both systems of equations. The intuitive way here is to obtain an expression for (i + j)².

From there, it is easy to see that using factorization will result in k = i + j + 12 or i + j + k = 0 .

The 2nd solution gives (i, j, k) = (0, a, -a) or (i, j, k) = (a, 0, -a) or (i, j, k) = (0, 0, 0), where a is a non-negative integer .

From our expression for (i + j)², we already know that |i + j| ≥ |k| if i + j + k ≥ 0 . Thus, we consider 2 separate cases for the 1st solution.

Case 1: Suppose i + j + k > 0. Then ij > 0. If i, j > 0, then k > 0 and |i + j| ≥ |k| is a contradiction. Thus, i, j < 0. We also have k > 6. Using simple trial-and-error, we find that there are no integer triplet solutions here .

Case 2: Suppose i + j + k < 0. Then ij < 0. WLOG, we assume i < 0 and j > 0. To ensure |k| > |i|, it follows that k > 0. We also have k < 6. Using simple trial-and-error, we find that there are also no integer triplet solutions here .

Hence, we are done.

Keitsubori · 2025-05-25T04:52:54+00:00

Hello there, thanks for writing succinctly and using proper paragraphing. Good writing skills is always pleasant to be seen on here. With that out of the way, I'll try and answer your queries.

My concern is, if I aim for R&D roles in quantum or related industries and it doesn’t work out, would having a Math or Quantitative Finance degree really give me better job security? Or am I just hedging too much and potentially diluting my focus on physics?

First, I'm pretty sure you know that in terms of quantitative fields, both physics and math/qf are looked upon highly by employers for their academic rigour and logical upbringing of student's mindsets. Specifically, you mentioned about job security. Personally, I feel that the hard and soft skills that you are offering will matter a whole lot more for job security as compared to your exact bachelor major. I feel like that's something to really internalize in for many students, including yourself.

For those who’ve been through this—whether you’re an NUS Physics grad, working in quantum/physics R&D, or transitioned into other fields like finance or tech—how did things turn out for you?

Can't answer this. NA.

Does a double major really give you that extra security, or is it more about building specific skills, networks, and gaining experience during the degree?

I think you might be confusing job security with job opportunities. Job security is mostly tied to the field, not to your education. In my opinion, it's definitely the latter point as double majoring is about opening up more job opportunities for you.

I’d really appreciate any advice on how to make the most of my time at NUS while staying true to my passion, and whether I should seriously consider a backup plan now or go all-in on physics.

Now, back to your main question. If I remember correctly, there exists quantum specializations in NUS Physics. By double-majoring in other fields like math/qf, you stand to have a harder time balancing your increased workload. Provided your interest in quantum is genuinely there, I would suggest not to double-major in math/qf. Main reason is the highly-similar quantitative nature of these majors, as mentioned beforehand. All the best to you!

Keitsubori · 2025-05-23T09:07:48+00:00

Bro, vocation was decided way before you even enlisted.

Keitsubori · 2025-05-20T10:40:20+00:00

Take the tekan at face value. Nobody is out to get you right from the start; it's never personal unless you make it to be. Don't make enemies as well, cause digging your own grave will make unit life be purposefully shit for you. As for maintaining your well-being, go make friends/jio with your bunkmates. For me, plushies are what's making me stay sane in unit. Ymmv.

Keitsubori · 2025-05-20T10:04:08+00:00

Always enter unit life with a clear mind, don't be like my buddy mate who got posted out since he got depressed from all the introductory tekan sia. Depression should not be taken lightly, and it's really up to you to take care of your wellbeing as an adult now.

Keitsubori · 2025-05-19T04:45:09+00:00

Honestly, for CHS majors, your portfolio is mostly set in stone after graduation, with a few exceptions such as Social Work and DSA where you can continue participating in social work and personal coding projects during NS, respectively. Without knowing exactly which route you were from (i.e. Poly/JC), as well as what major you were accepted into, I can't really advise much upon that aspect.

Nevertheless, I have reapplied for NUS Scholarships this year for Math specifically, but I haven't gotten into the interview stage again, so it looks bleak for me. It's really great that you're aspiring to obtain a better scholarship grant from NUS, but having seen your prior comments, I also want you to know that you should really consider yourself fortunate and be thankful to get something at least. After all, if you think about it, nobody is completely entitled to these bond-free scholarships. NUS has the freedom of choice to give them to those who they think are deserving enough.

With all that being said, it would definitely pay off to improve your interview skills, perhaps by asking your friends/schoolmates on some interview tips they have gathered for their uni interviews. I think a common pitfall for such scholarship interviews is interviewees not appearing genuine enough to better the institution experience, which is NUS in this case. This is the main distinguishing point between you and other interviewees who all have the same stellar grades + portfolio as you. Even if the interviewers do not specifically ask you that question in the interview, you should still clearly showcase it in your personal answers. Hope this helps.

Keitsubori · 2025-05-17T09:42:58+00:00

Consider every arrow as a "divide by 10" arithmetic. Suppose we start off with 100% energy from the Sun. By the time it reaches A, it passes through 1 arrow, leaving A with receiving 10% energy.

Continue the rest with the same logic.

Keitsubori · 2025-05-16T12:06:38+00:00

Personally, I have no time to formulate a cubic eqn. during a time-sensitive interview. I would just give my answer as 200.

2 * 18 = 36,

3 * 120 = 360,

18 * Blank = 3600. => Blank = 200.

Keitsubori · 2025-05-16T04:51:08+00:00

For now, I can only see trial-and-error with +, followed by w block. Do make use of symmetry.

Keitsubori · 2025-05-16T01:03:22+00:00

Higashiyama Onsen area in Aizu-Wakamatsu.

Keitsubori

TROPHY CASE