Math and CS book club: currently reading: 東京大学 理系 2020年度 第4問 解説

This is an unscheduled reading session of the math and CS book club approaching short math related web articles or article snippets.

Currently reading (June 16th - June 21st)

Upcoming readings

  • feel free to make suggestions :slight_smile:

Basic Vocab


How is the reading going? Any questions so far?

To gauge the level of Japanese and the length of the texts we want to read during the bi-weekly math related reading sessions let’s have some polls. In case you haven’t read the text but still want to poll please use the dedicated option :sweat_smile:

Level of Japanese
  • too easy
  • slightly too easy
  • ideal
  • slightly too difficult
  • too difficult
  • I haven’t read the text - just want to poll :upside_down_face:

0 voters

Length of the text
  • too short
  • slightly too short
  • ideal
  • slightly too long
  • too long
  • I haven’t read the text - just want to poll :upside_down_face:

0 voters

Japanese-wise? I think the level is pleasant to begin with, no problems there so far.

Maths-wise? I’m being stubborn and trying to do the proof myself but number theory has never been my strong suit…still wanna prove to myself that I could make it into 東大 though :no_mouth: I’m gonna give myself another day, then I’m gonna check out the solution.


Thanks a lot for sharing your thoughts. How did it go? I mean, welcome at tôdai :grin:

…there’s always a second chance for everything, right? :sweat_smile: I’m working on a list of useful vocab and phrases for my Anki deck though. That way I’ll at least master the problem language-wise :grin:

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I have a grammar question about a phrase in the solution:

ある自然数 a, m が a(a+1)=m^n を満たすとする

From context, I’d say the not-so-literal translation is “Assume there are natural numbers a and m such that a(a+1) = m^n.” When I try to do a more literal translation I get stuck at the first ある. What grammar point is this? I guessing the meaning is something like “there are natural numbers a and m” but I’m a bit confused by seeing ある at the beginning of a sentence.

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It’s this one: 或る -

which I would translate as “arbitrary but fixed” (I don’t know how you’d say that properly in English, this is me literally translating it from German ^^)


Ahh yes I know what you mean, thank you!

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Yes, that’s the spirit! And there is a whole list of problems we can choose from :blush:

Thanks for sharing the link :slight_smile:

Also thanks to those of you who voted. Looks like the level of Japanese and length of the text are convenient.

Although this thread is supposed to be bi-weekly I’d like to postpone for one more week to give our first CS related reading a chance. Therefore, next reading in this thread will start April 14th / 15th depending were you are based.

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Found the thread again and slightly updated the description. :sweat_smile:

For those who are interested let’s have a look into mondai four from last years entrance exam: 東京大学 理系 2020年度 第4問 解説.

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This looks interesting, I shall have to check it out tomorrow!

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Hey! So I just found the home thread for this and I am really excited for the upcoming book club starting on June 30th. Thanks @tls for setting this up! It’s a really cool idea to combine some other interests (math and CS) with learning Japanese

In the end soooo much went over my head haha (more than like 70%). Like oftentimes I could understand the meaning of everything in the sentence but just not get my mind around what the thing as a whole was referring to. Honestly even if this was in English I would still have to reread it multiple (multiple, multiple) times to get what it’s saying and I know I would still not understand a very significant of it. But still this was really interesting and it was really fun trying to comprehend everything

Some things/comments...

When they say 足し合わせる, they are referring to addition right? If so, why not just use 足す?

How have I never seen ∏ before?

Is this an easier or harder problem than most Tokyo University math entrance exam questions? I… really hope this is a harder than average question.

Interesting that より can also be used as “from”

:exploding_head: I think if I was able to follow the Japanese more my mind would melt a little bit less when I read the math parts… Like to a certain extent, this kind of makes sense to me from what the original problem says, but to every other extent this would take me a few hours to sort out :crazy_face:

≦ or ≤? I wonder why they prefer the former? Is the first one more common in Japan?


Thanks for sharing. I usually experienced that exercises get more easy once I spent more time with the topic in general. Because I have the routine in going through them and I have all common definitions and theorems ready for access. But you powered through it… and in Japanese :muscle:

Awesome. Only one more week and we will start :slight_smile:

I think 足す is used in expressions like “What is the result when you add two and two?” whereas 足し合わせる is used in expressions like “To calculate the mean of a series of n numbers you first sum them up and then divide this by n”.

Yes, depending on what your field of interest is the \Sigma might be more present.

In case you are interested: one topic the \Pi is used in would be relational agebra which is used to provide the theoretical foundation for relational databases. Another one would be the geometric mean.

I don’t know about that but you can look into it a little bit yourself if you like. On the website they have tasks from different years.

I don’t know either. @AWR san or @Naphthalene senpai any thoughts to share? (Maybe as well on the 足す vs 足し合わせる? :slight_smile: )

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The former is the de facto standard way to write it in Japan.
Here’s a random blog post about it: 意外と知らない『≦』と『≤』 - 株式会社IPOMOEA 技術翻訳イポメア 公式ブログ | 意外と知らない『≦』と『≤』

What do you want to know?

Edit: Ah, I just saw the original message. 足す is to add (+), 足し合わせる is to do a summation, to sum up ( ∑ or just multiple + in a row).


I fully admit that I cannot resist working on any interesting math problems I come across, this was a fun one. I solved 1 and 2 pretty quickly, I thought 2 was an especially nice problem, but 3 took me a lot longer. Now I need to go back and actually look up all that vocab, because I initially went through the problem just using the equations to understand what it was asking. (I used to do a lot of competitive math)

Are these are proof questions? A fun challenge would be learning how to write math proofs in Japanese.


Thank you @tls and @Naphthalene for your responses they were really helpful! Some good to know information when I want to look over other entrance exam questions in the future.

Wow!!! This is really impressive especially without looking at the Japanese text portions!


I’ve been thinking the same, it would be nice to try and get used to that! But with this problem I’m still stuck trying to solve parts 2b) and 3) in any language, let alone in Japanese :sweat_smile:


One small thing I noticed here is that 以上 (and presumable 以下 as well) are inclusive comparisons, so in a math context 2 以上 is \geq2 not >2.

English version of my proof for part (1)
\begin{align*}(2^n-1)(2^n-1) &= \left(\sum_{k=0}^{n-1}2^k\right)^2\\ &=\sum_{0\le j,k\le n-1}2^j\cdot2^k\\ &=2\cdot\left(\sum_{0\le j<k\le n-1}2^j\cdot2^k\right)+\sum_{k=0}^{n-1}2^{2k}\\ &=2\cdot a_{n,2}+\sum_{k=0}^{n-1}2^{2k}\\ a_{n,2}&=\frac{(2^n-1)^2-\sum_{k=0}^{n-1}2^{2k}}{2} \end{align*}



(In line math doesn’t seem to work inside of a details thing)

\begin{align*} 3\cdot b_n&=\sum_{j=0}^n3\cdot2^{2j}\\ &=\sum_{j=0}^n\left(2\cdot2^{2j}+2^{2j}\right)\\ &=\sum_{j=0}^n\left(2^{2j+1}+2^{2j}\right)\\ &=\sum_{j=0}^{2n}2^j\\ &=2^{2n+2}-1\\ b_n&=\frac{2^{2n+2}-1}{3} \end{align*}

Substituting into our original equation,

\begin{align*} a_{n,2}&=\frac{(2^n-1)^2-\sum_{k=0}^{n-1}2^{2k}}{2}\\ &=\frac{(2^n-1)^2-b_{n-1}}{2}\\ &=\frac{(2^n-1)^2-\frac{2^{2n}-1}{3}}{2}\\ &=\frac{3(2^n-1)^2-2^{2n}+1}{6}\\ &=\frac{3\cdot2^{2n}-3*2^{n+1}+1-2^{2n}+1}{6}\\ &=\frac{2\cdot2^{2n}-6\cdot2^n+2}{6}\\ a_{n,2}&=\boxed{\frac{2^{2n}+2}{3}-2^n} \end{align*}

I wrote this with very few actual words, so hopefully making a Japanese version won’t be too bad, but my vocabulary and grammar knowledge in this area is severely lacking. There’s definitely a specific way of writing for math proofs in English, so I’m assuming Japanese is the same, with plenty of set phrases you wouldn’t see elsewhere. I am going to try to translate this based on some language used in the explanation, but I’m especially unsure about how to declare my definition for b. I’d welcome any suggestions on both Japanese tips and parts of the proof I should explain more. I tend to write very concise proofs, and it’s also been 6 or 7 years since I had to write a math proof for someone else to read.

For my own purposes, I also reworded the problem statement in English, involving literally no actual translation. (spoiled instead of collapsed so I can use inline math)

Let n,k be integers with 1\le k \le n. For a given n, choose k distinct


Then consider the sum of all _nC_k such selections, and call that a_{n,k}. For example,


(1) For n\ge2, what is a_{n,2}?

(2) For n\ge1, think about the following function of x


What are \frac{f_{n+1}(x)}{f_n(x)} and \frac{f_{n+1}(x)}{f_n(2x)} in terms of x?

(3) Express \frac{a_{n+1,k+1}}{a_{n,k}} in terms of n,k.

I also wrote a proof of part (2) consisting of only equations, but I would probably need to actually prove the first statement I make there.

Very brief part 2 proof
\begin{align*} f_n(x)&=\prod_{j=0}^{n-1}(1+2^jx)\\ f_{n+1}(x)&=\prod_{j=0}^{n}(1+2^jx)\\ \frac{f_{n+1}(x)}{f_n(x)}&=\boxed{1+2^nx}\\ f_{n}(2x)&=\prod_{j=0}^{n-1}(1+2^j\cdot2x)\\ &=\prod_{j=1}^{n}(1+2^jx)\\ \frac{f_{n+1}(x)}{f_n(2x)}&=\boxed{1+x} \end{align*}

I didn’t know this forum supported formulas, nice :no_mouth:

Based on this problem I think the set phrase would be "bを[…]とする。”