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Let x and y be two-digit integers such that y is obtained by reversing

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Let x and y be two-digit integers such that y is obtained by reversing  [#permalink]

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New post 25 Mar 2019, 00:12
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E

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  55% (hard)

Question Stats:

42% (02:09) correct 58% (02:20) wrong based on 19 sessions

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Re: Let x and y be two-digit integers such that y is obtained by reversing  [#permalink]

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New post 25 Mar 2019, 00:31
Bunuel wrote:
Let x and y be two-digit integers such that y is obtained by reversing the digits of x. The integers x and y satisfy x^2 - y^2 = m^2 for some positive integer m. What is x + y + m ?

(A) 88
(B) 112
(C) 116
(D) 144
(E) 154



Let, x = ab = 10a+b
i.e. y = ba = 10b+a

\(x^2 - y^2 = (x+y)*(x-y) = 11(a+b)*9(a-b) = m^2\)

i.e. \(99(a^2-b^2) = m^2\)

Since 9 is a perfect square so \(11(a^2-b^2)\) should also be a perfect square i.e. \((a^2-b^2) = (a+b)*(a-b)\) must be 11

i.e. \(a+b = 11\) and \(a-b = 1\)

i.e. \(m^2 = 3^2*11^2\)

i.e. \(m = 33\)

i.e. x+y+m = 11(a+b)+m = 11*11+33 = 154

Answer:Option E
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Let x and y be two-digit integers such that y is obtained by reversing  [#permalink]

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New post 25 Mar 2019, 00:49
Bunuel wrote:
Let x and y be two-digit integers such that y is obtained by reversing the digits of x. The integers x and y satisfy x^2 - y^2 = m^2 for some positive integer m. What is x + y + m ?

(A) 88
(B) 112
(C) 116
(D) 144
(E) 154


Since x and y are both two-digit integers and y is obtained by reversing
the digits of x, if y = 10a + b, then the value of x is 10b + a

\(x^2 - y^2 = 100a^2 + b^2 + 20ab - 100b^2 - a^2 - 20ab = 99(a^2 - b^2)\)

\(x^2 - y^2 = m^2 = 99(a+b)(a-b) = 3*3*11*(a+b)*(a-b)\)

If a = 6 and b = 5, then we have a perfect square for \(m^2\) and m =33

Therefore, the value of x + y + m = 65 + 56 + 33 = 154(Option E)
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Let x and y be two-digit integers such that y is obtained by reversing  [#permalink]

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New post 28 Mar 2019, 23:33
given x = 10a+b , y = 10b+a
given \(x^2-y^2 = m^2\)

11(a+b) 9(a-b) = \(m^2\)
\(99(a^2-b^2) = m^2\)

\(9*11(a^2-b^2) = m^2\)

\((a^2-b^2)\) should be equal to 11 or 11 *4 which should be a perfect square .

\(a^2-b^2\) so considering it as 11. ( we can consider it as 11*4 a perfect square , but if we consider it would be above the answer/options range )

(a+b)(a-b) = 11*1=> a = 6 , b =5 , m = 3*11 = 33

x+y+m = 11(a+b) + m = 11(11)+33 = 154 ( Option E )
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Let x and y be two-digit integers such that y is obtained by reversing   [#permalink] 28 Mar 2019, 23:33
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