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04 Dec 2019, 02:12
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E
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Difficulty:
95%
(hard)
Question Stats:
53% (03:08) correct
47%
(03:28)
wrong
based on 128
sessions
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Not Attempted Yet
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
(A) 88
(B) 112
(C) 116
(D) 144
(E) 154
Are You Up For the Challenge: 700 Level Questions
Archit3110
04 Dec 2019, 02:56
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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
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
Ansh777
Joined: 03 Nov 2019
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GMAT 1: 710 Q50 V36
Schools: Booth '24 Columbia CBS '24 Darden '24 Fuqua '24 Haas '24 Harvard '24 LBS '24 Stanford '24 Kellogg '24 Wharton '24
GMAT 1: 710 Q50 V36
Posts: 53
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04 Dec 2019, 05:43
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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
Answer : E
Posted from my mobile device
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
Answer : E
Posted from my mobile device
