How many six-digit integers have the sum of their two leftmost digits equal to the sum of their two rightmost digits?

In base-10, where the sum of two digits is less than 10, and where the sum of these digits are equal,  the number of different leftmost digits is one less than the number of different rightmost digits as the pair of digits with a leading zero cannot exist in the rightmost pair but can exist in the leftmost pair [e.g. 20xx02 and 20xx20, but not 02xx20 or 02xx02]:
n*(n+1) where n = 1 to 9.
(1*2)+(2*3)+(3*4)+(4*5)+(5*6)+(6*7)+(7*8)+(8*9)+(9*10) = 330.

In base-10, where the sum of the two digits is greater than 10 to the maximum total of 18, there is an equal number of different leftmost and rightmost pair of digits, with the number of different pairs decrementing by 1 as the sum increments. In other words: n² where n = 1 to 9. 
(1*1)+(2*2)+(3*3+(4*4)+(5*5)+(6*6)+(7*7)+(8*8)+(9*9) = 285.

There exists 100 different pairs for the middle two digits: 00 to 99.

Therefore there are 61,500 different six-digit numbers where the leftmost pair of digits is the same sum as the rightmost pair of digits.
(330+285)*100 = 61500

Edited on March 4, 2012, 5:26 pm

Posted by Dej Mar on 2012-03-04 12:28:37