Could you have tested into Japan’s First Higher School?

Japanese
#1

I can’t remember whether I’ve mentioned this at Bunpro or not, but the main way I currently use Japanese is to translate college entrance exams. Being a math professor by day, I’m trying to use these problems in my classes to build my students’ flexibility and deductive reasoning.

There are several sites I like, but one in particular has a huge database by school and year. Here is the oldest test I can find on the site: the First Higher School’s exam from June 1887. The original test is here.

Here’s my translation. Most of the Japanese is quite archaic, so I had to lean on Google Translate more than I’m proud of. But you can see a few Bunpro grammar items in the original text. I thought it was pretty cool seeing kanji/ateji that’ve fallen into disuse and verbs ending in ふ. Please feel free to correct me, but I did try to err on the side of clarity to an English-speaking mathematician and trying not to make the translation literal.

This test is really hard, especially considering pocket calculators wouldn’t exist for decades. This school eventually became part of Tokyo University, and 東大 is the hardest college in Japan to get into.

Oh, one more thing: I don’t have an answer key or anything. Enjoy!

Computation

  1. (a) Express these numbers as Roman numerals: 24; 365; 2547.
    (b) Use short division to divide 103420 by 4, then by 5, then by 6, then by 7.
  2. State the rule for finding the greatest common divisor for three or more numbers and explain the reasoning in detail. (You may assume that we already know the rule and how it works two numbers. That does not need to be included.)
  3. When calculating costs for publishing a certain book, the letterpress fee for one volume is 288 yen, 40 sen, the newspaper subscription fee is 187 yen, 20 sen, the printing fee is 93 yen, 15 sen, the book-binding fee is 162 yen, and advertising and miscellaneous expenses amount to 69 yen, 25 sen. Find the desired list price if we sell the book for 80% of that list price and have a profit equivalent to 50% of the total capital.
  4. (a) The current calendar year has 97 leap years for every 400 years. Find the average length of a year in days, hours, minutes, and seconds.
    (b) Consider the difference between the above average and 365 days, 5 hours, 48 minutes, and 46 seconds. How many years would it take of adding that difference for to amount to a full day?
  5. (a) A 13-meter ladder is flush with a vertical wall. If we move the bottom of the ladder away from the wall so that the top of the ladder has descended by 1 meter, how far is the bottom of the ladder from the wall?
    (b) Express that length in shaku (尺).

Algebra

  1. Simplify the following two expressions.
    (a) [𝑥/(1+𝑥)+(1–𝑥)/𝑥]÷[𝑥/(1+𝑥)-(1–𝑥)/𝑥]
    (b) 𝑥/{𝑦+𝑥/[𝑦+𝑥/(𝑦+𝑥/𝑦)]}
  2. Find 𝑎 so that 𝑥³+a𝑥–30 is divisible by 𝑥–5, with no remainder.
  3. State the rules for fraction addition and explain the reasoning behind them.
  4. Solve this equation.
    (4𝑥+3)/9+(7𝑥–29)/(5𝑥–12)=(8𝑥+19)/18.
  5. The sum of a number and some part of it is equal to 𝑛 times the difference between the original number and that part. Then what fraction of the original number does this added/subtracted part represent?

Geometry

  1. Prove that the sum of any two sides of a triangle is greater than the length of the third side.
  2. Prove that for any right triangle ABC, the sum of the areas of the squares drawn on legs AB and AC is the area of a square drawn on hypotenuse BC (without using the principle of proportionality).
  3. Explain the meaning of each of the words below.
    (a) Plane.
    (b) Circle.
    (c) Circumference.
  4. Prove that opposite angles of a cyclic quadrilateral are supplementary.
  5. On triangle ABC, for midpoint D between vertex A and side BC, prove that when AD is half the length of BC, angle BAC is right.
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#2

Here’s my valiant effort on the rabbit geometry paper :rabbit2:

  1. :small_red_triangle: + :small_red_triangle_down: > :triangular_ruler:

  2. :leg: :red_square: + :leg: :blue_square: = :hippopotamus: :purple_square:

  3. (a) :airplane:
    (b) :red_circle:
    (c) :large_blue_circle: ferrets

Gotta admit that 4 and 5 foxed me though :fox_face:

#3
  1. a) XXIV; CCCLXV; MMDXLVII
    b) 100E3/4 + 34E2/4 + 20/4 = 25E3+8,5E2+5=25855
    10E4/5 + 30E2/5+40E1/4+20/5=[…]=20684
    10E4/6 → carry 4 +43E3/6 → carry 1+13E2/6 → carry 1 +12E1/6 → carry 0 = 17236 +4/6
    14777 + 2/7 yeah I got bored of explaining
  2. faster explaining it in Python:
    def gcd(a, b):
    while b != 0:
    a, b = b, a % b
    return a
    (Euclide algorithm) (a%b = the remain of the division of a/b)

letterpress_fee+sub_fee+print_fee+bind_fee+ad_expenses= a
0.8x=1.5a
find x
(x= 1481.25)
4. a)
(365303+9797)/400
too lazy to do the rest let’s assume it’s 365,2 (pi=3 btw)
b)86400 s in a day
my result - the result on the paper = delta
86400/ delta = result (in years)
5. a)square triangle → pythagora babyyyy
a2+b2=c2
a=(c2-b2)^1/2
a=(169-144)^1/2= 25^ 1/2 = 5
b) time for a break
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#4
  1. Simplify the following two expressions.
    (a) [𝑥/(1+𝑥)+(1–𝑥)/𝑥]÷[𝑥/(1+𝑥)-(1–𝑥)/𝑥]

get nominator and denominator on the same common denominator : (1+x)*x.
I’ll do it for the nominator to prove my big brain power
(x/1+x)*x/x + (1-x)/x *(1+x)/(1+x)= (x2/(1+x)*x)+(1-x)(1+x)/((1+x)*x) identite remarquable → (a-b)(a+b) = a2-b2
= (x2+1-x2)/((1+x)x)
now same for the denominator, too lazy to write down on keyboard again. with the common denominator you simplify the expression and you get 1/(2x^2-1)

(b) 𝑥/{𝑦+𝑥/[𝑦+𝑥/(𝑦+𝑥/𝑦)]}

start from the inner most fraction and then the second inner most etc… and solve, it takes time and it’s a pain to write it down here so assume that your big brain can find the result with no issue, at the end make sure to multiply nominator with denominator to get a simplify result, might end up with y^4 though (4 layers)

  1. Find 𝑎 so that 𝑥³+a𝑥–30 is divisible by 𝑥–5, with no remainder.

basically solve f(5)=0 no ? it’s a=-19

  1. State the rules for fraction addition and explain the reasoning behind them.

put same denominator then add. how can they can ask this after all the lengthy fraction we had to simplify this breaks my mind ???

  1. Solve this equation.
    (4𝑥+3)/9+(7𝑥–29)/(5𝑥–12)=(8𝑥+19)/18.

got bored of lengthy equations, skipping this for now

  1. The sum of a number and some part of it is equal to 𝑛 times the difference between the original number and that part. Then what fraction of the original number does this added/subtracted part represent?

x+a=n*(x-a) ?
a= xn-na-x
I want to get rid of the a on one side
a+na = xn-x
a(1+n)=xn-x
a=(xn-x)/(1+n)
I want to isolate x
a=(x(n-1)/(1+n)
a/x = (n-1)/(1+n)

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#5

I only ever got a handful of passing grades for math in the first year of high school. After that, I failed all math classes until the end of school.

Despite that, I’m sure I could have gotten into Japan’s First Higher School. As a janitor.

#6
  1. Prove that the sum of any two sides of a triangle is greater than the length of the third side.

ABC triangle
coordonnees
A(x1,y1)
B(x2,y2)
C(x3,y3)
AB =sqrt(sq(x1-x2)+sq(y1-y2))
AC=sqrt(sq(x1-x3)+sq(y1-y3))
BC=(sqrt(sq(x2-x3)+sq(y2-y3))

triangle inequality : AB + BC = AC if colinear
then in the case of a triangle
AB+BC > AC
not sure if this would be tolerated tho

  1. Prove that for any right triangle ABC, the sum of the areas of the squares drawn on legs AB and AC is the area of a square drawn on hypotenuse BC (without using the principle of proportionality).
  2. Explain the meaning of each of the words below.
    (a) Plane.

All of his points can be identified within a planar system constitued by 2 unparalleled lines

(b) Circle.

All of his points can be identified within the distance r of its radius and the angle theta ((r,θ); r appartient a R et theta appartient a [0;2pi]

(c) Circumference.

considering the previous point, 2pi*r

  1. Prove that opposite angles of a cyclic quadrilateral are supplementary.

bless you

  1. On triangle ABC, for midpoint D between vertex A and side BC, prove that when AD is half the length of BC, angle BAC is right.

so we need to prove pythagora was right
ABC triangle 2AB+2BC=2*AC
using some formula we need to get back on ABD triangle and prove that the resulting formula is pythagora therefore BAC is right ?

okay I found another way:
in any triangle the midpoint of a segment is the center of the circle that passes through the endpoints of the segment.
In a right triangle, the right angle lies on the circle whose diameter is the hypothenuse (merci Thales mon khoya)
A lies on the circle with diameter BC, the radius of this circle is AD; conclusion : if a triangle has a median to a side equal to half that side, the triangle must be right angled at the opposite vertex.

this or using coordinates
B(0,0)
C(c,0)
A(x,y)
D(c/2,0)
express the expression of the circle with D as its center and a c/2 radius,
calculate sqAB + sqAC and find out it’s equal to sqBC, congrats you’ve calculated pythagora’s theorem

quite a lengthy test, I wonder how I would have performed at 14 on this, i’m twice this age and my brain relies too much on computer calculation lol

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#7

I REALLY HOPE YOU DON’T USE THE ALGEBRA PART IN A TEST TO YOUR STUDENTS YOU MONSTER

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