Take a two-digit number. The possible numbers for original number condition are 29, 38, 47, 56, 65, 74, 83, 92.
<u>SOLUTON:
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Given, Take a two-digit number. Write it backwards. Add it to the original number. If what you get is a perfect square,
We have to find what could be the original number? Find all answers
Now, let the original number be xy
Then, xy + yx is a perfect square.
⇒ (10x + y) + (10y + x) is perfect square
[ because we can write 21 as 2 x 10 + 1
]
⇒ (11x + 11y) is a perfect square
⇒ 11(x + y) is perfect square.
11(x + y) to be perfect square x + y should also be 11.
Then, x + y = 11.
Now, possible values of x and y are (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), (9, 1)
(10, 1) is not possible because x range is from 0 – 9 as we took it as digit in xy
Now, the originals numbers xy are 29, 38, 47, 56, 65, 74, 83, 92.
Hence, the possible numbers for original number condition are 29, 38, 47, 56, 65, 74, 83, 92.
