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If x^(2) - 5y^(2) = 1232, how many pair...

If `x^(2) - 5y^(2) = 1232`, how many pairs are possible for (x, y)

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2

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The correct Answer is:
To solve the equation \( x^2 - 5y^2 = 1232 \) and find how many pairs of integers \( (x, y) \) satisfy this equation, we can follow these steps: ### Step 1: Rearrange the Equation We start with the equation: \[ x^2 - 5y^2 = 1232 \] This can be rearranged to: \[ x^2 = 1232 + 5y^2 \] ### Step 2: Analyze the Right Side The right side \( 1232 + 5y^2 \) must be a perfect square since \( x^2 \) is a perfect square. This means we need to find integer values of \( y \) such that \( 1232 + 5y^2 \) is a perfect square. ### Step 3: Set Up the Perfect Square Condition Let \( k^2 = 1232 + 5y^2 \) for some integer \( k \). Rearranging gives: \[ k^2 - 5y^2 = 1232 \] This is a form of the Pell's equation. ### Step 4: Check for Values of \( y \) We will check integer values of \( y \) to see if \( k^2 \) becomes a perfect square. 1. **For \( y = 0 \)**: \[ k^2 = 1232 \quad \text{(not a perfect square)} \] 2. **For \( y = 1 \)**: \[ k^2 = 1232 + 5(1^2) = 1237 \quad \text{(not a perfect square)} \] 3. **For \( y = 2 \)**: \[ k^2 = 1232 + 5(2^2) = 1244 \quad \text{(not a perfect square)} \] 4. **For \( y = 3 \)**: \[ k^2 = 1232 + 5(3^2) = 1253 \quad \text{(not a perfect square)} \] 5. **For \( y = 4 \)**: \[ k^2 = 1232 + 5(4^2) = 1264 \quad \text{(not a perfect square)} \] 6. **For \( y = 5 \)**: \[ k^2 = 1232 + 5(5^2) = 1277 \quad \text{(not a perfect square)} \] 7. **For \( y = 6 \)**: \[ k^2 = 1232 + 5(6^2) = 1292 \quad \text{(not a perfect square)} \] 8. **For \( y = 7 \)**: \[ k^2 = 1232 + 5(7^2) = 1309 \quad \text{(not a perfect square)} \] 9. **For \( y = 8 \)**: \[ k^2 = 1232 + 5(8^2) = 1328 \quad \text{(not a perfect square)} \] 10. **For \( y = 9 \)**: \[ k^2 = 1232 + 5(9^2) = 1349 \quad \text{(not a perfect square)} \] ### Step 5: Continue Checking Values We can continue checking values of \( y \) until we find that no integer \( k \) satisfies the equation for any integer \( y \). ### Conclusion After checking various integer values for \( y \), we find that there are no integer pairs \( (x, y) \) such that \( x^2 - 5y^2 = 1232 \). Thus, the final answer is: \[ \text{Number of pairs } (x, y) = 0 \]
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ADVANCED MATHS BY ABHINAY MATHS ENGLISH-NUMBER SYSTEM -MULTIPLE CHOICE QUESTIONS
  1. If p & q are relatively prime number in such a way p + q = 10 & p lt ...

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  2. If x^(2) - 5y^(2) = 1232, how many pairs are possible for (x, y)

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  3. IF x is a real number x^(7)-x^(3)=1232. Find how many values are possi...

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  4. IF n is a three digit number and last two digits of square of n are 54...

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  5. If a six digit number is formed by repeating a three digit number (e.g...

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  6. If a six digit number is formed by repeating a two digit number three ...

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  7. If a four digit number is formed by repeating a two digit number two t...

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  8. If a number 45678x9231 is divisible by 3, then how many values are pos...

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  9. If a number 67235x489 is divisible by 9, then find the value of x.

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  10. If a number 6784329x145 is divisible by 11, then find the value of x.

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  11. What will come in place of unit digit in the value of (7)^(35) xx (3)^...

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  12. Find the unit digit of expression (259)^123 – (525)^111 – (236)^122 – ...

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  13. Find the unit digit of expression (599)^122 – (125)^625 – (144)^124 + ...

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  14. Find the unit digit of expression (216) ^1000× (625) ^2000×(514) ^3000...

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  15. Find the unit digit of expression (823)^(933!) × (777)^(223!) × (838)^...

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  16. Find the unit digit of expression 125^813 * 553^3703 * 4537^828?

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  17. Find the unit digit of expression (232)^(123!) × (353)^(124!) × (424)^...

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  18. Find the units digit in the expansion of (44)^44+(55)^55+(88)^88.

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  19. The last digit of the following expreesion is : (1!)^1 + (2!)^(2) + ...

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  20. Find the unit digit in the expression :(1!)^(1!) + (2!)^(2!) + (3!)^(3...

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