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

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Math Expert
Joined: 02 Sep 2009
Posts: 58453
Let x and y be two-digit integers such that y is obtained by reversing  [#permalink]

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25 Mar 2019, 00:12
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55% (hard)

Question Stats:

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

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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

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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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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

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

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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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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 )
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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