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Take any 3 digit number, difference between first digit and last digit should be greater than 1. Reverse the digits. Find the difference. Reverse the digits in this number (difference). Add them to get 1089 always.For example, let the number as 109, reversing the digits gives 901, difference between them 901 - 109 = 792, reversing the digits in the difference gives 297, adding them gives 792 + 297 = 1089.
This is not something that I discovered, I just know about this property through some source long time back. But I will attempt to show why this happens.
Let the random 3 digit number be abc. Its value is 100a + 10b + c
Reversing the digits gives cba. Its value 100c + 10b + a.
Difference between them assuming that a > c (for sake of simplicity of analysis)
|abc - cba| = abc - cba = 100a + 10b + c - (100c +10b + a) = 99 x (a-c)
Let, |abc - cba| = xyz (another 3 digit number)
xyz, then is a multiple of 99 i.e. of both 11 and 9.
Properties of multiples
- In multiples of 9, the sum of digits is divisible by 9
- In multiples of 11, the difference between sum of all odd and even digits is divisible by 11
In this case,
Since xyz is divisible by 9, x + y + z is divisible by 9 or x + y + z = 9 or 18. It can't be 27 or more.
Since xyz is divisible by 11, x + z - y = 0 or 11. It can't be 22 or more.
If x + z - y = 0, then x + z = y, then x + y + z = 2y
2y cannot be 9, it can be 18, so 2y = 18 or y = 9
If x + z - y = 0, then x + z = y or x + z = 9, this is possible
If x + z - y = 11, then x + z = 11 + y or x + z = 20, this is not possible
So, we can conclude that, y = 9 and x + z = 9.
Reversing the digits in xyz, we get zyx.
Adding the 2 numbers, xyz + zyx = 100x + 10y + z + 100z + 10y + x
Total = 101(x + z) + 20y
Substituting for x + z = 9 and y = 9, we get,
Total = 101 x 9 + 20 x 9 = 909 + 180 = 1089!
By the way, 33 square is 1089. 99 x 11 is 1089 and 121 x 9 is 1089.