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28 Nov 2023, 10:45
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
24% (02:43) correct
76%
(02:15)
wrong
based on 68
sessions
History
Date
Time
Result
Not Attempted Yet
How many positive integers n exist such that both 48n + 4 and 27n + 19 are perfect squares?
A. 0
B. 1
C. 2
D. 3
E. 4
A. 0
B. 1
C. 2
D. 3
E. 4
29 Nov 2023, 00:39
Kudos
Bookmarks
Bunuel
\(48n + 4 = x^2\) ⇒ \(n = \frac{x^2 - 4}{48}\)
\(27n + 19 = y^2\) ⇒ \(n = \frac{y^2 - 19}{27}\)
\(\frac{x^2 - 4}{48} = \frac{y^2 - 19}{27}\)
Multiplying both sides by \(3\)
\(\frac{x^2 - 4}{16} = \frac{y^2 - 19}{9}\)
\(9x^2 - (4*9) = 16y^2 - (16*19)\)
\((16*19) - (4*9) = 16y^2 - 9x^2\)
\( 16y^2 - 9x^2 =4 ((4*19) - (1*9))\)
\( (4y+3x)(4y-3x) =4 ((4*19) - (1*9))\)
\( (4y+3x)(4y-3x) =4*(67)\)
Case 1
\(4y + 3x = 67\)
\(4y - 3x = 04\)
Adding both the equations
\(8y = 71\)
The value of y results in a fraction. Hence, we can ignore this.
Case 2
\(4y + 3x = 134\)
\(4y - 3x = 02\)
Adding both the equations
\(8y = 136\)
\(y = 17\)
To find the value of \(x\), we can substitute the value of \(y\) in the equation \(4y - 3x = 02\)
\(x = 22\)
The value of \(n = 10\)
Case 3
\(4y + 3x = 268\)
\(4y - 3x = 01\)
Adding both the equations
\(8y = 269\)
The value of y results in a fraction. Hence, we can ignore this.
Hence, for only one value of \(n\) are both \(48n + 4\) and \(27n + 19\) perfect squares.
Option B
