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# Let p and q be two digit integers such that q is obtained by reversing

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Senior Manager
Joined: 13 Jun 2013
Posts: 266
Let p and q be two digit integers such that q is obtained by reversing  [#permalink]

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27 Jan 2015, 05:46
1
9
00:00

Difficulty:

75% (hard)

Question Stats:

63% (03:23) correct 37% (03:01) wrong based on 103 sessions

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Let p and q be two digit integers such that q is obtained by reversing the digits of p. The integers p and q satisfy the equation $$p^2 - q^2 = r^2$$ for some positive integer r. what is the value of p+q+r ?

A) 88
B) 112
C) 116
D) 144
E) 154
Math Expert
Joined: 02 Aug 2009
Posts: 7961
Re: Let p and q be two digit integers such that q is obtained by reversing  [#permalink]

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27 Jan 2015, 07:40
ans is E... the quickest method i can think of..
let the nos be10x+y and 10y+x..
so (10x+y)^2-(10y+x)^2=r^2..
we get 99(x-y)(x+y)=r^2.. now for x,y and r to be int, x and y can be 6 and 5 respectively..
r becomes 33 ... therefore p+q+r = 65+56+33=154..
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Joined: 03 Jul 2015
Posts: 30
Re: Let p and q be two digit integers such that q is obtained by reversing  [#permalink]

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18 Sep 2015, 06:11
chetan2u wrote:
ans is E... the quickest method i can think of..
let the nos be10x+y and 10y+x..
so (10x+y)^2-(10y+x)^2=r^2..
we get 99(x-y)(x+y)=r^2.. now for x,y and r to be int, x and y can be 6 and 5 respectively..
r becomes 33 ... therefore p+q+r = 65+56+33=154..

how do you exactly find the right choice 6,5 without knowing the range of selection? or did u randomly choice for the equation?
Manager
Joined: 29 Jul 2015
Posts: 155
Re: Let p and q be two digit integers such that q is obtained by reversing  [#permalink]

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18 Sep 2015, 13:06
1
manpreetsingh86 wrote:
Let p and q be two digit integers such that q is obtained by reversing the digits of p. The integers p and q satisfy the equation $$p^2 - q^2 = r^2$$ for some positive integer r. what is the value of p+q+r ?

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

let p = 10A+B
Then q = 10B+A

given that
$$p^2 - q^2 = r^2$$ for positive integer r.
plugging in values of p and q we get
$$(10A+B)^2-(10B+A)^2 = r^2$$
or $$100A^2 + B^2 + 20AB -100B^2 - A^2 - 20AB = r^2$$
or $$99A^2-99B^2=r^2$$
or $$9*11*(A+B)(A-B)=r^2$$ ......(1)

For r to be integer, $$\sqrt{r^2}$$ should be integer.
So, LHS of (1) should be perfect square.
9 is already a square. Either A+B or A-B should be equal to 11 to get square of 11.
A-B cannot be 11 as difference between 2 single digit numbers cannot be a two digit number.
So, A+B = 11 and A,B can be (6,5),(7,4),(8,3),(9,2)
Also A-B should be a perfect square. A-B can be equal to 1 or 4. If you see the values of A and B from above sets,A-B cannot be 4 in any case. So, A-B should be 1.
This is possible only when A = 6 and B = 5
plugging in these values of A and B in (1), we get r=33
Also, p=65 and q=56
So,
p+q+r= 65+56+33 = 154

Intern
Joined: 16 Aug 2015
Posts: 18
Let p and q be two digit integers such that q is obtained by reversing  [#permalink]

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18 Sep 2015, 14:26
Here is how i solved this problem :

P= 10*a + b ; Q = 10*b + a by combining the two we obtain : P+Q = 11*(a+b) and P-Q = 9(a-b).

Now we have that P^2−Q^2=r^2, this equation can be transformed into this one (P+Q)*(P-Q) = R^2 which implies that :

R^2 = 3^2 * 11* (a+b)*(a-b). For now all the job is done, all we need to observe is that r must be an integer, thus (a+b) must be equal to 11 and (a-b) must be a perfect square, which from all the possible combinations, the unique solution is a = 6 and b=5.

Now that we have all our inputs, we can convert all the abstract p and q and ... and.. to real life numbers 65 and 56 and r is 33 so the sum of the three numbers equal to 154. so answer (E) is correct!

Hope it helps
Math Expert
Joined: 02 Aug 2009
Posts: 7961
Re: Let p and q be two digit integers such that q is obtained by reversing  [#permalink]

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19 Sep 2015, 03:53
anik19890 wrote:
chetan2u wrote:
ans is E... the quickest method i can think of..
let the nos be10x+y and 10y+x..
so (10x+y)^2-(10y+x)^2=r^2..
we get 99(x-y)(x+y)=r^2.. now for x,y and r to be int, x and y can be 6 and 5 respectively..
r becomes 33 ... therefore p+q+r = 65+56+33=154..

how do you exactly find the right choice 6,5 without knowing the range of selection? or did u randomly choice for the equation?

Hi,
99(x-y)(x+y)=r^2..
now 99 is 9*11..
so x+y has to be 11, or multiple of 11..
since x and y are single digit, it has to be 11.... x and y can be 9,2;8,3;7,4;6,5...
you can easily see only 5 and 6 give us r as perfect square...
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Re: Let p and q be two digit integers such that q is obtained by reversing  [#permalink]

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20 Oct 2018, 21:34
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Re: Let p and q be two digit integers such that q is obtained by reversing   [#permalink] 20 Oct 2018, 21:34
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