Department of Mathematics Education

Dr. J. Wilson, EMAT 6690

by David Wise

**Problem**

A palindrome is a number that reads the same backwards as forwards, such as 2662 or 9,351,359. Prove that every six-digit palindrome is divisible by 11. Make a conjecture about any palindrome that has an even number of digits and prove it.

**Proof**

In order to prove that every six-digit palindrome is divisible
by 11, we must first discuss the divisibility rule of 11. Consider
any number, n, written in base 10. Let k be the number of digits
in n and D_{i} be the "i"th digit, from left
to right, in the number, where i = 1 to k. This number can be
expressed in the summation:

*note: * is used as the multiplication symbol*

n = (D_{k} * 10^{k-1}) + (D_{k-1} *
10^{k-2}) + ... + (D_{2} * 10^{1}) + (D_{1}
* 10^{0}). (1)

Therefore, 11 divides n if and only if 11 divides (D_{k}
* 10^{k-1}) + (D_{k-1} * 10^{k-2}) + ...
+ (D_{2} * 10^{1}) + (D_{1} * 10^{0}).

Modular arithmetic states that the number n is divisible by 11 if expression (1) is also divisible by 11 after the base 10 is replaced by the remainder of 10 divided by 11. This remainder can be expressed by (10 mod 11). Thus, 11 divides n if and only if:

11 divides (D_{k} * (10 mod 11)^{k-1}) + ...
+ (D_{2} * (10 mod 11)^{1}) + (D_{1} *
(10 mod 11)^{0}). (2)

Expression (2) can be simplified because 10 mod 11 is equal to –1. Therefore, 11 divides n if and only if:

11 divides (D_{k} * (-1)^{k-1}) + ... + (D_{2}
* (-1)^{1}) + (D_{1} * (-1)^{0}). (3)

-1 raised to any even power is equal to 1 and –1 raised to any odd power is equal to –1, so expression (3) becomes a summation of the digits involving alternated signs. 11 divides n if and only if:

11 divides (- D_{k} + ...
- D_{2} + D_{1}) for
n with an even number of digits, and

11 divides (D_{k} + ... -
D_{2} + D_{1}) for n with an odd number of digits.

This is known as the alternating summation rule for divisibility by 11. There are two versions of the alternating summation rule. One version states that a number is divisible by 11 if the alternating sum of the digits is also divisible 11, meaning that the alternating sum must be a multiple of 11. The second version states that a number is divisible by 11 if the sum of every other digit starting with the right-most digit is equal to the sum of every other digit starting with the second digit from the right, meaning that the alternating sum is 0.

The second version of the alternating summation rule will be the most convenient and useful method to prove that every six digit palindrome is divisible by 11. Consider any six digit palindrome, n = abccba, where a, b, and c are digits of n and a, b, and c can be equal. We calculate the alternating sum: -a + b – c + c – b + a = 0. Therefore, every six digit palindrome is divisible by 11.

In fact, I claim that every even digit palindrome is divisible by 11. Consider any even digit palindrome, n = ab…ba, where a, b, … are digits of n and a, b, … can be equal. All even digit palindromes are "evenly" symmetric, meaning that each digit has a "matching partner." Let’s illustrate with a few examples: aa, abba, abccba, abcddcba, abcdeedcba, etc. In each case, and for all even digit palindromes, the line of symmetry divides the palindrome with an even number of digits on each side. Applying the second version of the alternating summation rule, we find that the alternating sum of the digits will always equal 0. In general, -a + b … – b + a = 0. Therefore, every even digit palindrome is divisible by 11.

Mulliss, C. Why do the divisibility ‘rules’ work?
*Math Forum*. Retrieved April 19, 2000 from the World Wide
Web: **http://forum.swarthmore.edu/k12/mathtips/mulliss.html**.

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