This question asks for the remainder when the sum of two numbers raised to a large power is divided by another number. We can solve this efficiently using modular arithmetic.
Understanding the Problem
We need to find the remainder of the expression $2023^{2024} + 2025^{2024}$ when divided by $2024$. In mathematical terms, we want to calculate:
$$ (2023^{2024} + 2025^{2024}) \pmod{2024} $$Applying Modular Arithmetic
Let's consider the divisor, which is $2024$. We can express the numbers $2023$ and $2025$ in terms of $2024$ using modular arithmetic:
- $2023$ is one less than $2024$. So, $2023 \equiv -1 \pmod{2024}$.
- $2025$ is one more than $2024$. So, $2025 \equiv 1 \pmod{2024}$.
Now, we can substitute these equivalences into the original expression:
$$ (2023^{2024} + 2025^{2024}) \pmod{2024} \equiv ((-1)^{2024} + 1^{2024}) \pmod{2024} $$Calculating the Powers
Next, we need to calculate the powers:
- $(-1)^{2024}$: Since the exponent $2024$ is an even number, raising $-1$ to an even power results in $1$. So, $(-1)^{2024} = 1$.
- $1^{2024}$: Raising $1$ to any power always results in $1$. So, $1^{2024} = 1$.
Finding the Final Remainder
Substitute the calculated powers back into the expression:
$$ (1 + 1) \pmod{2024} $$Adding the results, we get:
$$ 2 \pmod{2024} $$The remainder when $2$ is divided by $2024$ is simply $2$. Therefore, the remainder of the original expression $2023^{2024} + 2025^{2024}$ when divided by $2024$ is $2$.
Conclusion
The remainder is 2.
Important Questions from Algebra (Notes)
In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If she/he attempts all 60 questions and secures 130 marks, the number of questions she/he attempts wrongly, are?
Match List-I with List-II
List-1 List-II (A) If $\begin{bmatrix}\lambda-1 & 0 \\ 0 & \lambda-1 \end{bmatrix} $, then $\lambda$ is (I) 0 (B) If A=$ \begin{bmatrix}1 & 2 \\2 & 4 \end{bmatrix} $, then $\Delta$ is (II) 1 (C) If A = $ \begin{bmatrix}1 & 0 \\0 & \frac{1}{2} \end{bmatrix} $, then $|A^{-1}|$ is (III) -2 (D) If $ \begin{bmatrix}a+1 & 1 \\1 & 2 \end{bmatrix} = \begin{bmatrix}-1 & 1 \\1 & 2 \end{bmatrix} $, then a is (IV) 2 Choose the correct answer from the options given below:
- If (x - 1) is a factor of $2x^2 - 5x + k = 0$, then the value of k is:
- If $x = (2+\sqrt{3})^{\frac{1}{3}} + (2+\sqrt{3})^{-\frac{1}{3}}$ and $x^3-3x + k = 0$, then the value of k is:
If \[ \begin{vmatrix} \dfrac{1}{a(b+c)} & \dfrac{1}{b(c+a)} & \dfrac{1}{c(a+b)} \\ \dfrac{bc}{a(b+c)} & \dfrac{ca}{b(c+a)} & \dfrac{ab}{c(a+b)} \\ 1 & 1 & 1 \end{vmatrix} = k, \text{ then the value of } k \text{ is:} \]
