How do you prove that any palindrome, a number that reads the same backwards and forwards, with an even number of digits is divisible by 11?

I know how to show a number is divisible by 11, but how do you prove the above case?

How do we know a number is divisible by 11?

e.g. is 164587467 divisible by 11?

Add up the odd-number positioned digits
1+4+8+4+7 = 24
6+5+7+6 = 24
if their sum has a difference of zero or a multiple of 11, then the original number divides by 11

e.g. 16437003
sum of even-number positioned digits = 6+3+0+3 = 12
sum of odd-number positioned digits = 1+4+7+0 = 12

now forming the palindrome would obviously make the original odd-number positioned numbers into the even-number positioned digits and vice versa.

so their sum would still have either a difference of zero or a multiple of 11.

BTW, the same property would result if you formed the palindrome of a number with an even number of digits

e.g. 1737032 and 2307371 are palidromes, and both are divisible by 11

the sum of the digits of their respective groupings are 6 and 17

and the difference between these two is 11

To prove that any palindrome with an even number of digits is divisible by 11, you can follow these steps:

Step 1: Consider an arbitrary palindrome with an even number of digits. Let's use the example of the palindrome "abccba". Here "a" and "b" represent the digits.

Step 2: Rewrite the palindrome as a sum of terms. Each term would represent a digit multiplied by an appropriate power of 10. In our example, "abccba" can be written as:

abccba = a * (10^5) + b * (10^4) + c * (10^3) + c * (10^2) + b * (10^1) + a * (10^0)

Step 3: Observe that the digit pattern in our sum alternates in sign: positive, negative, positive, negative, and so on. This pattern occurs because the digits are being multiplied by either 1 or -1.

Step 4: Rearrange the terms to group the positive and negative terms separately:

abccba = (a * (10^5) - b * (10^4) + c * (10^3) - c * (10^2) + b * (10^1) - a * (10^0))

Step 5: Factor out 11 from each pair of positive and negative terms:

abccba = a * (10^5 - 1) - b * (10^4 - 1) + c * (10^3 - 1) - c * (10^2 - 1) + b * (10^1 - 1) - a * (10^0 - 1)
= a * (99999) - b * (9999) + c * (999) - c * (99) + b * (9) - a * (1)

Step 6: Simplify further and notice that each term is divisible by 11:

abccba = a * (99999) - b * (9999) + c * (999) - c * (99) + b * (9) - a * (1)
= 99999a - 9999b + 999c - 99c + 9b - a
= (99999a - a) + (9b - 9999b) + (999c - 99c)
= 99990a - 9900b + 900c
= 11(9090a - 900b + 80c)

Step 7: Since the final expression is a multiple of 11, we can conclude that any palindrome with an even number of digits is divisible by 11.

In summary, the proof shows that for any palindrome with an even number of digits, you can write it as a sum of terms alternating in sign. By grouping these terms and factoring out 11, it becomes evident that the sum is divisible by 11.