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작성자	인생의고통
작성일	2011-03-18 21:35:58 KST	조회	2,440
제목	중학교 2학년 수학 문제

Solution:

This problem can be solved simply with modular arithmetic:

a = 6 mod 14; b = 1 mod 14; x = r mod 14.

x^2 - 2ax + b = 0 is equivalent to

x^2 - 12x + 1 = 0 mod 14

x(x - 12) = -1 mod 14

-1 has no proper divisors modulo 14, thus factors x and x-12 must correspond to either 1 or -1 mod 14.

If x = 1 mod 14, then x-12 = 3 mod 14: false. If x = -1 mod 14, then x - 12 = 1 mod 14: true.

Therefore, x = -1 mod 14 = 13 mod 14, i.e., the remainder r of x is 13.

Second Solution:

Homura used in Episode 1 a basic approach with usual integer arithmetic.

Let x = 14q + r, a = 14s + 6, b = 14t + 1.

Substitute into x^2 - 2ax + b = 0 to get after some calculations

x^2 - 2ax + b = 14c + (r+1)^2 = 0 with c = 14q^2 + 2q - 28qs - 2rs - 12q + t - r

14 divides 14c and 0, hence it also divides (r+1)^2. This implies that r+1 is divisible by 14.

The question asks for a remainder r between 0 and 13, so we obtain r = 13.

중3선행문제로 추정

2번째문제

Solution:

By Fermat's Little Theorem, for any prime p and integer a,

a^p = a (mod p)

Thus:

(1+n)^p - n^p - 1 (mod p) is equivalent to

(1+n) - n - 1 (mod p) is equivalent to

0 (mod p)

So the overall expression is divisible by p.

Second solution:

The problem can be solved with the binomial theorem:

For a, b not equal to 0 and nonnegative integer p it holds that:

where the binomial coefficient (p choose k) is the integer

for 0 ≤ k ≤p. Therefore,

and

since (p choose 0) = (p choose p) = 1.

It holds that (p choose 1) = p. Since p is a prime number, the factor p in (p choose k) is not divisible by k, k-1,...,2 for 2 ≤ k ≤ p-1. Therefore, p divides (p choose k) for 1 ≤ k ≤ p-1.

Since each summand on the right side is divisible by p, the whole sum, i.e. the left side is also divisible by p.

시그마가 나오네요 고등학교 2 학년

문제 3번째 문제

Simplying the fraction.

Reference that for any variables a & b, (a-b) (a+b) = a^2-b^2. Let (2x + 1)^.5 = a and (2x - 1)^.5 = b.

Multiply F(x) by 1 or (a-b)/(a-b) or ((2x + 1)^.5 - (2x - 1) ^.5) / ((2x + 1)^.5 - (2x - 1)^.5). The

denomina;tor becomes:

((2x + 1)^.5 + (2x - 1)^.5) * ((2x + 1)^.5 - (2x - 1) ^.5)

= (2x + 1) - (2x - 1)

= 2

The numerator becomes:

(4x + (4x^2 - 1)^.5 ) ((2x + 1)^.5 - (2x - 1)^.5)

=(4x + ((2x)^2 - 1)^.5 ) (a - b)

=(4x + (2x - 1)^.5 (2x + 1)^.5 ) (a -b)

=(4x + a b ) (a - b)

=4x (a - b) + (a^2) b - a (b^2)

=4x (a - b) + (2x + 1) b - a (2x - 1)

=4x (a - b) - 2x (a - b) + a + b

=2x ( a - b) + a + b

=(2x + 1) a - (2x - 1) b

=a^3 - b^3

Therefore:

F(x) = (a^3 - b^3 ) / 2

Note that when x= 1, 2, 3, 4, ..., 60;

a^3 = 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5, 121^1.5;

b^3 = 1^1.5, 3^1.5, 5^1.5, 7^1.5, 9^1.5, ..., 119^1.5;

When summating over F(x) for x between 1 and 60, observe that the majority of the terms in a^3 and

b^3 cancel out, leaving:

(121^1.5 - 1^1.5)/2 =((11*11)^1.5 - 1)/2 = 665

Thus, sum of F(x) for x between 1 and 60 is 665. The solution can be generalized as equal to (a

(x_max)^3 - b(x_min)^3)/2.

What The Hell .....

Problem
A number sequence {F(n)} that can be defined as F(1) = 1, F(2) = 1, F(n+2) = F(n) + F(n+1) (where n is any natural number) is called the Fibonacci sequence and its general solution is given by,

<위 Problem 의역해석>

유리함수에서 제2영역인데 미지수가 2안팎임. 결국 무한4차계수로 들어가야함... 만약 무한4차계수로 했을경우 풀이법이

이렇게변합니다. ↓

Answer the following questions by using this fact if needed:

Define a sequence of natural numbers X(n) (where n is any natural number), in which each digit is either 0 or 1, set by the following rules:

(i) X(1) = 1

(ii) We define X(n+1) as a natural number, which can be gotten by replacing the digits of X(n) with 1 if the digit is 0, and with 10 if the digit is 1.

For example, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ...

Anecdote: X(n) can be considered a clever analogy to the show, where Episode 10 should replaces Episode 1 and Episode 1 replaces Episode 0 in the viewer's next viewing.

(1) Find A(n), defined as the number of digits of X(n)?

(2) Find B(n),defined as the numbers of times '01' appears in X(n)? For example, B(1) = 0, B(2) = 0, B(3) = 1, B(4) = 1, B(5) = 3,...

Solution
Part 1
Let A(n) equal to the number of digits in X(n), which consists solely of 1s and 0s. Let's suppose x(n) is the number of 0s in X(n) at n iteration (poor choice of variable by the student). Let's suppose y(n) is the number of 1s in X(n) at n iteration, then x(n) + y(n) = A(n). Since

Every time a 0 appears, it is replaced with 1 at the next iteration, contributing to a single 1 in X(n+1).
Every time a 1 appears, it is replaced with 10 at t he next iteration, contributing to a single 1 and a single 0 in X(n+1).
it follows that the number of 0s in the next iteration is equal to the number of 1s previously:

x(n +1)= y(n) ;

and the number of 1s in the next iteration is equal to the number of 0s AND the numbers of 1s previously:

y(n+1)=x(n)+y(n) .

Next, prove that x(n) is a fibonacci sequence, since we know that:

y(n+1) = x (n) + y(n) ;
x(n+2) = y(n+1) ;
x(n+1) = y(n) ;
by substition we can show,

x(n+2) = y(n+1) = x(n) + y(n) = x(n) + x(n+1)

x(n+2) = x(n+1) + x(n)

x(n) fits the definition of a fibonacci sequence. Since x(n) is a fibonacci sequence, it follows that y(n) is also a fibonacci sequence (since x(n+1) = y(n)). Therefore:

x(n+2)=x(n +1)+x(n)

y(n+2)=y(n+1)+y(n)

x(n+2)+y(n+2)=x(n+1)+y(n+1)+x(n)+y(n)

Recall: A(n)=x(n)+y(n), therefore

A(n+2)=A(n+1)+A(n) .

Since, X(1) = 1, X(2) = 10, X(3) = 101, X(4) = 10110, X(5) = 10110101, ...

It follows that A(1) = 1, A(2) = 2, A(3) = 3, A(4) = 5, A(5) = 8,...

Recall fibonacci sequence F(n)= {1, 1, 2, 3, 5, 8, ...}. Therefore A(n) = F(n+1), or

Part 2
Let B(n) be the number of times '01' appears in X(n). Any two digits in X(n) may be 00, 01, 10, 11, and the corresponding digits in the next iteration X(n+1) will be

00 -> 11
01 - > 110
10 -> 101
11 -> 1010
Thus, any two digit sequence in X(n) that begins with 1* will contribute to a 01 sequence in the next iteration. In other words, B(n+1) equals y(n), except when unit digit of X(n) is 1, or:

B(n +1) = y(n) - odd(X(n))

where

odd(n) = 1 if n is odd (unit digit is 1)
odd(n) = 0 if n is even (unit digit is 0)
y(n) is a fibonacci sequence with starting values: y(1) = 1, y(2) = 1, y(3) = 2 thus y(n) = F(n)

B(n+1) = F(n) - odd(X(n))

B(n) = F(n-1) - odd(X(n-1)), or