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Math Expert V
Joined: 02 Sep 2009
Posts: 60778
Let p and q be two digit integers such that q is obtained by reversing  [#permalink]

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1
8 00:00

Difficulty:   65% (hard)

Question Stats: 50% (03:02) correct 50% (03:33) wrong based on 20 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

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GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 5751
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: Let p and q be two digit integers such that q is obtained by reversing  [#permalink]

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1
p=10a+b
q=10b+a
$$p^2 - q^2 = r^2$$
we get
(11a+11b)(9a-9b)=r^2
r= 3*√11*(a+b)*(a-b)
a+b has to be 11 ; 6+5 ;
so r = 33
no ; 65+56+33 ; 154
IMO E

Bunuel 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

Are You Up For the Challenge: 700 Level Questions
Manager  S
Joined: 03 Nov 2019
Posts: 54
Re: Let p and q be two digit integers such that q is obtained by reversing  [#permalink]

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p= 10A+B ; q=10B+A

Also A and B are first digit of 2 digit numbers so they cannot be zero hence A and B lies between 1 and 9.

p+q=11*(A+B)

p-q=9*(A-B) ........................................consider all the four statement above as eq (i)

r^2= p^2-q^2 = (p+q)(p-q) =99(A+B)(A-B)

r is a positive integer therefore r cannot be 0 or A is not equal to B and A>B

and maximum possible value of A-B would be 9-1 =8................................consider above two statement above as eq (iii)

Now for 99(A+B)(A-B) should be a square term

so minimum possible value of (A+B)(A-B)=11

And since 11 is prime so either (A+B) =11 or (A-B)=11

But from iii A+B = 11 therefore A-B= 1

thus r is a multiple of 11*3=33

r = 33, 66, 99, 132, 165, 198 ....=33K ...K=some constant

Minimum value of

p+q+r= 11(A+B)+33 .............(ii)

=11*11+33= 121+33 = 154

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VP  P
Joined: 24 Nov 2016
Posts: 1143
Location: United States
Re: Let p and q be two digit integers such that q is obtained by reversing  [#permalink]

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Bunuel 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

p=10a+b, q=10b+a
r^2=p^2-q^2, r^2=(p+q)(p-q)
r^2=(10a+b+(10b+a))(10a+b-(10b+a))
r^2=(11a+11b)(9a-9b), r^2=11*9*(a+b)(a-b)
r^2 = positive integer, (a+b)(a-b)>0, 9≥a>b>0

9 = perf square
11*(a+b)(a-b) = perf square
(a+b) = multiple of 11 because 8≥(a-b)≥0
(a,b) = [9,2; 8,3; 7,4; 6,5]
(a-b) = perf square
(a,b) = [6,5]

r^2=11*9*11*1=33
p=65, q=56, r=33, p+q+r=154

Ans (E) Re: Let p and q be two digit integers such that q is obtained by reversing   [#permalink] 06 Dec 2019, 07:59
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# Let p and q be two digit integers such that q is obtained by reversing  