This article deals with questions 1-5. (Out of 7 questions)
University level / general level. According to the Japan Mathematical Examination Association, the pass rate is 5.7%, and all are descriptive. It is often said that it is more difficult than the entrance examination questions of the University of Tokyo.
I will try to solve some of the Math Test Level 1 Primary Exam Sample Questions. (The model answer is available here [https://www.su-gaku.net/digitalbook/past_question/1q_ans_1ji/index_h5.html#1].
Python......paiza.io FORTRAN ...... ABC187 code test in AtCoder code test
Question 1 Find the minimum value of the positive integer n that satisfies 2018n ≡ 2 mod (1000).
Will be. First of all, what is a mod?
What is mod When p ≡ q (mod n), p-q is divisible by n.
Did you understand somehow? Let's look at an example.
Example 3 ≡ 8(mod 5) 3-8 is divisible by 5.
It's a symbol like this.
I will solve this by full search (algorithm that tries all patterns). (Actually, the first digit is used for classification> the second digit is used for classification.)
Python
n = 1
while True:
if (2018*n-2)%1000 == 0:
break
n += 1
print(n)
389
FORTRAN
program question1
do n = 1,1000
!Since it is mod 1000 here, we can prove that n does not take a value greater than 1001.
if (mod((2018*n-2),1000) == 0) then
print '(i0)', n
end if
end do
end
389
889
Therefore, the smallest 389 is considered to be the answer (although 1000 and 2018 are even numbers from the beginning, so a full search up to 500 is fine).
Swift
var n = 0
while n < 1001{
if (2018*n-2)%1000 == 0{
print(n)
break
}
n += 1
}
It is an implementation in a while statement.
389
When I saw the answer, it was correct.
Find tan (2 Arctan (1/3) + Arctan (1/12)). However, Arctan x represents the inverse function of tan x, where -π/2 <Arctan x <π/2.
What is "Arctan is the inverse function of tan"?
See this this article in detail.
I would like to calculate this normally and find a close value.
import math
inside = 2*math.degrees(math.atan(1/3))#Contents of tan
inside += math.degrees(math.atan(1/12))
print(math.tan(math.radians(inside)))
0.8888888888888887
Is it about 8/9 (correct answer)? I wonder if I really use Arctan's addition theorem ...
Policy
--Let's realize Arctan's addition theorem
――Decimal numbers are prone to error, so let's think in fractions
--Let's create a function that satisfies the above two conditions.
Wikipedia says:
\arctan u+\arctan v=\arctan \left({\frac {u+v}{1-uv}}\right){\pmod \pi },\qquad uv\neq 1\,.
Here, since it is defined only by -π/2 to π/2, (mod π) is omitted. Think of it as u = a/b, v = c/d. In the problem, after that, it is restored with tan, so only (u + v)/(1-uv) needs to be considered (tan (arctan (x)) = x). After all, the answer is (u + v)/(1-uv), so consider a program that outputs this.
For the time being, I made a function like this.
from fractions import Fraction
def arctan_sum(a,b,c,d):
u = Fraction(a,b)
v = Fraction(c,d)
return ((u+v),(1-u*v))
The return value is a tuple. If you write as below,
from fractions import Fraction
def arctan_sum(a,b,c,d):
u = Fraction(a,b)
v = Fraction(c,d)
return ((u+v),(1-u*v))
left = arctan_sum(1,3,1,3)
total = arctan_sum(left[0],left[1],1,12)
8/9
I got a beautiful answer.
A tetrahedron with four points (1, -4,1), (2,2,2), (2, -6, -3), (3, -2, -1) in the xyz space as vertices Find the volume.
Solve using Salas's formula. First, create a function to find the volume of a tetrahedron whose vertices are (0,0,0) (x1, y1, z1), (x2, y2, z2), (x3, y3, z3). However, since we have to consider a tetrahedron with (0,0,0) as the apex, we would like to calculate x1 ~ z3 by hand (because it seems to be the fastest).
Python
from fractions import Fraction
def sarasu(x1,y1,z1,x2,y2,z2,x3,y3,z3):
return Fraction(abs(y1*z2*x3+z1*x2*y3+x1*y2*z3-z1*y2*x3-x1*z2*y3-y1*x2*z3),6)
from fractions import Fraction
def sarasu(x1,y1,z1,x2,y2,z2,x3,y3,z3):
return Fraction(abs(y1*z2*x3+z1*x2*y3+x1*y2*z3-z1*y2*x3-x1*z2*y3-y1*x2*z3),6)
print(sarasu(1,6,1,1,-2,-4,2,2,-2))
3
This was also a simple answer. (It was the correct answer.)
3rd order square matrix A= \left(
\begin{array}{ccc}
4 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{array}
\right)Answer the following questions about.
① Find the eigenvalue.
Python You can find the eigenvalues in Python, so I will ask for it.
import numpy as np
import numpy.linalg as LA
a = np.array([[4,1,1],[1,2,1],[1,1,2]])#Matrix for which you want to find eigenvalues
aeig_val = LA.eig(a)[0]
aeig_val = np.sort(aeig_val)
print(aeig_val)
[1. 2. 5.]
When I tried to list it, there was an error, so I made this result. (The answer is 1,2,5, so there is.)
I changed it to the form of the answer with replace. Python
import numpy as np
import numpy.linalg as LA
a = np.array([[4,1,1],[1,2,1],[1,1,2]])#Matrix for which you want to find eigenvalues
aeig_val = LA.eig(a)[0]
aeig_val = np.sort(aeig_val)
ans = str(aeig_val)
ans = ans.replace("[","").replace(".]","").replace(". ",", ")
print(ans)
1, 2, 5
Solving the eigenvalue problem numerically --I would like to find it by referring to Kochi University, Matrix power, 3 Identity matrix, Transpose of matrix (matrix). All you have to do is do this.
Python
import numpy as np
import numpy.linalg as LA
A = np.array([[4,1,1],[1,2,1],[1,1,2]])
I = np.array([[1,0,0],[0,1,0],[0,0,1]])
A3 = np.linalg.matrix_power(A, 3)#A^3
A2 = np.linalg.matrix_power(A, 2)#A^2
ans = A3-8*A2+18*A-12*I#Calculate with the operator as it is
print(ans)
[[2 1 1]
[1 0 1]
[1 1 0]]
If you write it in an easy-to-understand manner,
\left(
\begin{array}{ccc}
2 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}
\right)
(This was also the correct answer)
It is known that the number of machine failures within a certain period follows the Poisson distribution. One company uses two machines A and B, and when machine A follows an average of 2.5 failures and machine B averages 4.5 times in a year, answer the following questions. However, assuming that the number of failures of A and B in one year is independent of each other, the Poisson distribution table at 1-6-6 is used for the answer, and the answer is rounded off to the fourth decimal place and the third decimal place. Ask for.
① Find the probability that Machine A will fail more than 4 times a year.
The Poisson distribution is ...
"The probability function that represents the Poisson distribution is
P(k) = (e^{-k}*λ^k) / k!
is. In other words, the probability that a random event that occurs an average of λ times per unit time occurs k times per unit time is P (k). "
(Excerpt from Beautiful story of high school mathematics: Meaning and mean / variance of Poisson distribution)
In this problem, the probability of being broken less than 4 times is subtracted from the whole. At this time, it can be seen that λ is 2.5 and k is 0 to 3. Python
zero_three = 0.0821+0.2052+0.2565+0.2138#Just added
print(round((1-zero_three)*1000)/1000)
0.242
It was the correct answer.
It is known that the number of machine failures within a certain period follows the Poisson distribution. One company uses two machines A and B, and when machine A follows an average of 2.5 failures and machine B averages 4.5 times in a year, answer the following questions. However, assuming that the number of failures of A and B in one year is independent of each other, the Poisson distribution table at 1-6-6 is used for the answer, and the answer is rounded off to the fourth decimal place and the third decimal place. Ask for.
② Find the probability that machines A and B will fail 8 times or more in total.
First, list the values of the Poisson distribution with an average of 2.5 times and 4.5 times. Next, find the probability that the number of failures will be less than 8, and subtract it from the whole. Python
A = [0.0821,0.2052,0.2565,0.2138,0.1336,0.0668,0.0278,0.0099,0.0031,0.0009,0.0002]#Average 2.5 Poisson distribution
B = [0.0111,0.0500,0.1125,0.1687,0.1898,0.1708,0.1281,0.0824,0.0463,0.0232,0.0104]#Average 4.5 Poisson distribution
ans = 0
for i in range(8):
for j in range(i+1):
ans += A[i-j]*B[j]
print(round((1-ans)*1000)/1000)
0.401
It was the correct answer.
** Pass (5 questions on the pass line)! ** However, there is a problem that it is faster to calculate normally than to program.
-Qiita Markdown Writing Summary -Mathematics test, Arithmetic test homepage -Introduction to Japan NAG Fortran -Question about Yahoo Answers error -Tech Academy Iterative processing! How to use while/for statements written in Swift [for beginners] -Calculate trigonometric functions in Python (sin, cos, tan, arcsin, arccos, arctan) -Wikipedia Inverse Trigonometric Function -fractions --- rational numbers -LaTeX command collection matrix -Sort based on arbitrary row / column with NumPy sort and argsort functions --Solving the eigenvalue problem numerically --Kochi University -Matrix power -3 Identity matrix, matrix transpose (matrix) -Rounding decimals and integers in Python round and Decimal.quantize
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