@x2maverickc said: Nahi bhai! ye bhi nahi, @YouMadFellow ise dekhna How many 4 digit nos. have the property that the sum of the thousand and hundred place digits is equal to the sum of ten's and unit digit ? The new PG interface presents us with a lot of difficulty to post something in bold. May it happen now!!!
Edited: I realized that I was very lazy..
Is it 2280 ? WRONG
I would have probably left it during the exam :(.. My approach is too lengthy..
Possible sums of two digits are from 0 to 18. That means say the number is abcd, a+b lies between 0 and 18 and so does c+d.
I looked into the possible numbers for each of these, and I saw a pattern coming up.
For 0, no numbers
For 1 ( 10 ) -> 2 possible numbers ( 1010, 1001 )
For 2 ( 20, 11 ) - > 6 possible numbers ( 2020, 2002, 2011, 1120, 1102 , 1111 )
For 3 ( 30, 21 ) -> 12 possible numbers (3030, 3003, 2121, 2112, 1212, 1221, trust me

)
So, the series is 2 + 6 + 12 + 20 + 30 +... but
After the 9th term, it starts decreasing.
And a new series crops up, and that is of Tn = n^2
Till 9th term, Sum = 330
For 9 (90,81,72,63,54) -> 90 possible numbers
For 10(91,82,73,64,55) -> 81 possible numbers ( every pair can be taken in 2 ways except 55)
For 18 (99) -> Only 1 possible number ( 9999 )
So, after 9th term, series is 9^2 + 8^2 + 7^2 + ... 1^2 = 9.10.19/6 = 285
Now, Total sum = 515 !!!
May be the destination is not for me, but the journey is..http://www.pagalguy.com/forums/chit-chat/shoutbox-t-83788/p-3594016/r-4343099