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The difference between two reversed 2-digit positive integers is a per

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New post 22 Jul 2019, 13:03
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

25% (02:47) correct 75% (02:06) wrong based on 24 sessions

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The difference between two reversed 2-digit positive integers is a perfect square. How many such pairs of integers are there?

A. 8
B. 9
C. 11
D. 12
E. 13
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New post 22 Jul 2019, 13:16
(12,21), (32,23), (43,34), (45,54), (56,65), (67,76), (78,87), (89,98): the difference between each of these pairs is 9.
(95,59), (84,48), (73,37),(62,26) , (51,15): the difference between each of these pairs is 36
Option (E)

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Re: The difference between two reversed 2-digit positive integers is a per  [#permalink]

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New post 22 Jul 2019, 13:51
1
ab-ba=x^2.(ab, ba -two digit integer)
10a+b-(10b+a)=x^2
10a+b-10b-a=9a-9b=9*(a-b)=x^2

What number do we need to multiply with 9 to get a square of number??
1) a-b=1(9*1=3^2)
2) a-b=4(9*4=6^2)

(a-b=9(9*9=9^2) is also possible, but to get 81, we need 3 digit numbers)

Case1: a-b=1
2-1=1
3-2=1
4-3=1
5-4=1
6-5=1
7-6=1
8-7=1
9-8=1
There are 8 different combinations.

Case2: a-b=4
5-1=4
6-2=4
7-3=4
8-4=4
9-5=4
There are 5 different combinations.

In total, there are 13 different combinations.
Answer choice is E.

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Re: The difference between two reversed 2-digit positive integers is a per   [#permalink] 22 Jul 2019, 13:51
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