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HOW TO PROVE THAT : 49+56(N^2 +1) CAN NEVER BE …
Let’s simplify the expression: \[ 49 + 56(n^2 + 1) = 49 + 56n^2 + 56 = 56n^2 + 105 \] We need to prove that: \[ k^2 \ne 56n^2 + 105 \quad \text{for any integer } k \text{ and } n \in \mathbb{N}_0 \] Proof by Contradiction: Assume there exists some \( n \in \mathbb{N}_0 \) and \( k \in \mathbb{Z} \)Read more
Let’s simplify the expression:
\[
49 + 56(n^2 + 1) = 49 + 56n^2 + 56 = 56n^2 + 105
\]
We need to prove that:
\[
k^2 \ne 56n^2 + 105 \quad \text{for any integer } k \text{ and } n \in \mathbb{N}_0
\]
Proof by Contradiction:
Assume there exists some \( n \in \mathbb{N}_0 \) and \( k \in \mathbb{Z} \) such that:
\[
k^2 = 56n^2 + 105
\]
Rewriting:
\[
k^2 – 56n^2 = 105
\]
This is a Diophantine equation of the form:
\[
k^2 – 56n^2 = 105
\]
It resembles a generalized Pell’s equation, but unlike standard Pell’s equations, this has a non-zero right-hand side.
To find integer solutions, test small values of \( n \):
– \( n = 0 \Rightarrow k^2 = 105 \) → not a perfect square
– \( n = 1 \Rightarrow k^2 = 56 + 105 = 161 \) → not a perfect square
– \( n = 2 \Rightarrow k^2 = 224 + 105 = 329 \) → not a perfect square
– \( n = 3 \Rightarrow k^2 = 504 + 105 = 609 \) → not a perfect square
– \( n = 4 \Rightarrow k^2 = 896 + 105 = 1001 \) → not a perfect square
– \( n = 5 \Rightarrow k^2 = 1400 + 105 = 1505 \) → not a perfect square
– \( n = 6 \Rightarrow k^2 = 2016 + 105 = 2121 \) → not a perfect square
And so on. No value of \( k^2 = 56n^2 + 105 \) becomes a perfect square for any non-negative integer \( n \).
Also note:
For \( k^2 \equiv 56n^2 + 105 \pmod{8} \), since:
\[
56n^2 \equiv 0 \pmod{8}, \quad \Rightarrow k^2 \equiv 105 \equiv 1 \pmod{8}
\]
Only \( k \equiv 1, 3, 5, 7 \pmod{8} \) will work. However, checking modulo 7:
\[
56n^2 + 105 \equiv 0n^2 + 0 = 0 \pmod{7}
\Rightarrow k^2 \equiv 0 \pmod{7}
\Rightarrow k \equiv 0 \pmod{7}
\]
So \( k = 7m \), and we get:
\[
(7m)^2 = 56n^2 + 105 \Rightarrow 49m^2 = 56n^2 + 105
\Rightarrow 7m^2 = 8n^2 + 15
\]
Now check modulo 7:
\[
8n^2 + 15 \equiv m^2 \pmod{7}
\Rightarrow (8n^2 + 15) \mod 7
\]
But trying all \( n = 0 \) to \( 6 \), none of the RHS becomes a multiple of 7 ⇒ contradiction.
Conclusion:
\[
See less\boxed{49 + 56(n^2 + 1) \text{ is never a perfect square for any } n \in \mathbb{N}_0}
\]
KINDLY GIVE ANSWER ALONG WITH PROOF
Problem: There are 1000 doors, all initially closed. 1000 people walk by these doors. The first person opens every door. The second person toggles every second door (i.e., closes doors 2, 4, 6, etc.). The third person toggles every third door (i.e., doors 3, 6, 9, etc.), and so on. The 1000th personRead more
Problem:
There are 1000 doors, all initially closed. 1000 people walk by these doors. The first person opens every door. The second person toggles every second door (i.e., closes doors 2, 4, 6, etc.). The third person toggles every third door (i.e., doors 3, 6, 9, etc.), and so on. The 1000th person only toggles door 1000. After all the people have passed, how many doors remain open?
Step 1: Understand the pattern
Each person toggles the state of doors that are multiples of their own number. For example, person 12 will toggle doors 12, 24, 36, and so on. The state of each door will change every time it is toggled.
Step 2: Think about how many times each door is toggled
Take any door number, say door 6. It gets toggled by person 1 (since 1 divides 6), person 2 (2 divides 6), person 3 (3 divides 6), and person 6 (6 divides itself). In general, a door will be toggled once for each of its positive divisors.
Step 3: Determine when a door ends up open
Since all doors start closed, each toggle changes its state. So, if a door is toggled an even number of times, it will end up closed. If it is toggled an odd number of times, it will end up open.
Step 4: Identify which door numbers are toggled an odd number of times
From number theory, we know that most numbers have an even number of divisors because divisors usually come in pairs (like 2 and 3 for 6, since 2×3 = 6). However, perfect squares have an odd number of divisors. For example, 36 has divisors: 1, 2, 3, 4, 6, 9, 12, 18, and 36. Notice that 6×6 = 36, so the factor 6 appears only once, not in a pair. This gives it an odd number of total divisors.
Therefore, only the doors with perfect square numbers will remain open at the end.
Step 5: Count the perfect squares between 1 and 1000
The perfect squares less than or equal to 1000 are:
1² = 1,
2² = 4,
3² = 9,
…
31² = 961.
So, there are 31 perfect squares between 1 and 1000.
Final Answer:
See less31 doors will remain open.
Which is the smallest continent by land area?
The smallest continent by land area is Australia. Key Details: Land Area: Approximately 8.6 million square kilometers (3.3 million square miles). Features: It is the flattest and driest inhabited continent. Often referred to as the "island continent" because it is surrounded by water. Includes mainlRead more
The smallest continent by land area is Australia.
Key Details:
Despite being the smallest continent, Australia is home to unique biodiversity, vast deserts, and vibrant cities.
See lessWhat was the main reason behind abandoning the article 370 in India?
The primary reasons behind the abrogation of Article 370 in India were political, economic, and social, aimed at integrating Jammu and Kashmir (J&K) more closely with the rest of the country. Here are the main reasons: 1. Full Integration of Jammu and Kashmir Article 370 provided J&K with spRead more
The primary reasons behind the abrogation of Article 370 in India were political, economic, and social, aimed at integrating Jammu and Kashmir (J&K) more closely with the rest of the country. Here are the main reasons:
1. Full Integration of Jammu and Kashmir
2. Curbing Terrorism and Separatism
3. Development and Economic Growth
4. Ensuring Equal Rights
5. Addressing Internal Security
6. Uniformity in Governance
The decision to abrogate Article 370 on August 5, 2019, was accompanied by the bifurcation of the state of Jammu and Kashmir into two Union Territories: Jammu and Kashmir, and Ladakh. While the move was supported by many for its long-term vision of integration and development, it also faced criticism for bypassing local consent and imposing changes under contentious circumstances.
See less