Explain the sum a two digit number and the number obtained by reversing its digit is a perfect square number how many such number exists
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Two digit number is expressed as: 10a + b
Revering the digits, it becomes: 10b + a
Sum = 10a + b + 10b + a = 11a + 11b
Sum = 11(a+b)
sum has to be a perfect square, which means (a+b) should be equal to 11, because 11*11 is the only possibility to give a perfect square number.
Hence the sum of digits should be 11
Possibilities: (2,9) , (3,8) , (4,7) , (5,6), (6,5) , (7,4) , (8,3) , (9,2)
29,38,47,56,65,74,83,92 are the two digit numbers whose sum with the respective digits reversed give a perfect square number.
Hope it helps.
Revering the digits, it becomes: 10b + a
Sum = 10a + b + 10b + a = 11a + 11b
Sum = 11(a+b)
sum has to be a perfect square, which means (a+b) should be equal to 11, because 11*11 is the only possibility to give a perfect square number.
Hence the sum of digits should be 11
Possibilities: (2,9) , (3,8) , (4,7) , (5,6), (6,5) , (7,4) , (8,3) , (9,2)
29,38,47,56,65,74,83,92 are the two digit numbers whose sum with the respective digits reversed give a perfect square number.
Hope it helps.
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the sum of a two digit number and the number obtained by reversing its digits is a square number How many such numbers are there
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