Question

13 and 31 are strange pairs of numbers such that their squares are169 and 961 are also mirror images of each other. Find two more such pairs. ​

Answer 1

SOLUTION

First let's explore 2 -digit possibilities.

Let the number have digits x &y

so number are: 10x +y & 10y+x

Now,

$= > (10x + y) {}^{2} = 100 {x}^{2} + {y}^{2} + 20xy = 100( {x}^{2} ) + 10(2xy) + {y}^{2} \\ = > (10y + x) {}^{2} = 100y {}^{2} + {x}^{2} + 20xy = 100( {y}^{2} ) + 10(2xy) + {x}^{2}$

Let the numbers obtained will be of three digits whose general representation:

100a +10b +c (where a,b,c are the three digit of the number.)

Comparing the square of above numbers we can see that the three digits number obtained by squaring have following digits: x^2, 2xy, y^2

so, condition we need are:

x^2, y^2<10 and 2xy<10 and so, xy< 5.

Therefore only such possibilities are:

=) (1,1), (1,2) (1,3)

Therefore other two pairs are:

(11,11) & (12,21)

hope it helps ☺️⬆️