Posted by **Eric** on Tuesday, April 10, 2012 at 8:41pm.

It was proven during the previous lecture that

1+2+3+....+n=[n(n+1)]/2.

Using that result, find the sum of all 3-digit palindromes: a palindrome is a number that can be written as ABA, where 1=<A<=9 and 0<=B<=9. Hint: decompose ABA = 100*A + 10*B + A and use above proven result.

- Discrete Mathematics -
**MathMate**, Tuesday, April 10, 2012 at 10:56pm
Note that A can vary from 1 to 9, while B can take on the digits 0 to 9.

Sum of first and third digits:

So for each value of A=a(i), the sum is 100a(i)+a(i)=101a(i).

For the 9 values of A, we have a sum of

101Σi for i=1 to 9.

Since each value of a(i) can be matched to 10 values of B, the above sum has to be multiplied by 10 to give

1010Σi for i=1 to 9.

Sum of the second digit (B)

Each digit of B is in the 10 position, so we need to multiply by 10.

The sum of the digits 0-9 are repeated 9 times for each value of A, so the sum of digits B is

9*10Σi for i=0 to 9

=90Σi for i=0 to 9

Finally, add up the two sums:

1010Σi for i=1 to 9.

+

90Σi for i=0 to 9

to get the total

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