Exploring the Chinese Remainder Theorem for Polynomials: CMI B.Sc. Entrance 2016- Subjective Problem 5

In today's discussion, we delve into a fascinating problem from the 2016 CMI B.Sc. entrance exam that draws on key concepts from number theory, specifically the Chinese Remainder Theorem (CRT), but applies them in the context of polynomials. The problem asks us to find a polynomial $P(x)$ that satisfies two conditions:

This setup mirrors the CRT for integers but applies it to the algebraic framework of polynomials.

The video walks through the similarities between integer and polynomial division and emphasizes how techniques like the Euclidean algorithm can be extended to polynomials. Using polynomial differentiation and integration, we solve the given conditions, ultimately arriving at a general form for $P(x)$ by adjusting constants.

A key takeaway is the parallel between solving congruences for integers using the Euclidean algorithm and doing the same for polynomials, underscoring the algebraic unity between these two domains.

Watch the Video

Key Findings and Explanations:

$$
P(x) \equiv 1 \quad\left(\bmod x^{100}\right) \quad \text { and } \quad P(x) \equiv 2 \quad\left(\bmod (x-2)^3\right)
$$

This highlights the deep connection between the two fields.

$$
P^{\prime}(x) \text { must be divisible by } x^{99} \text { and } \quad(x-2)^2
$$

$$
P(x)=a \cdot \frac{x^{101}}{101}-4 \cdot \frac{x^{100}}{100}+4 \cdot \frac{x^{99}}{99}+b
$$

$$
P_1(x) \cdot x^{100}+P_2(x) \cdot(x-2)^3=1
$$

Problem on Natural Numbers | TIFR B 2010 | Problem 4

Try this problem of TIFR GS-2010 using your concepts of number theory and congruence based on natural numbers.

Problem on Natural Numbers | TIFR 201O| PART B | PROBLEM 4


Which of the following statements is false?

  • There exists a natural number which when divided by $3$ leaves remainder $1$ and when divided by $4$ leaves remainder $0$
  • There exists a natural number which when divided by $6$ leaves remainder $2$ and when divided by $9$ leaves remainder $1$
  • There exists a natural number which when divided by $7$ leaves remainder $1$ and when divided by $11$ leaves remainder $3$
  • There exists a natural number which when divided by $12$ leaves remainder $7$ and when divided by $8$ leaves remainder $3$

Key Concepts


NUMBER THEORY

CONGRUENCE

CHINESE REMAINDER THEOREM

Check the Answer


Answer:There exists a natural number which when divided by $6$ leaves remainder $2$ and when divided by $9$ leaves remainder $1$

TIFR 2010|PART B |PROBLEM 12

ELEMENTARY NUMBER THEORY DAVID M.BURTON

Try with Hints


Let us take the equations $x\equiv1(mod 3)$ and $x\equiv0(mod 4)$

Now we will apply Chinese remainder theorem to get the value of $x$

Since $3$,$4$ are relatively prime,gcd($3$,$4$)$=1$. Let $m=3\times4=12$

Then $M_1=4$,$M_2=3$.

Then gcd($M_1$,$3$)$=1$,gcd($M_2$,$4$)$=1$

Since gcd($4$,$3$)$=1$,therefore the linear congruence equation $4x\equiv1(mod 3)$ has a unique solution and $x\equiv1(mod 3)$ is the solution.

Since gcd($3$,$4$)$=1$,therefore the linear congruence equation $3x\equiv0(mod 4)$ has a solution and $x\equiv4(mod 4)$ is the solution.

Therefore,$x=1\times4\times1 +0\times3\times4=4$ is a solution.

The solution of the given system is $x\equiv4(mod 12)$

So we have used the Chinese Remainder Theorem to check the statements, you may use it to check for other options.

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