A palindrome is a word, phrase, or positive integer that reads the same forwards and backwards. For example, "kayak" is a palindrome. The integers \(292\), \(11\), and \(6\,357\,536\) are also palindromes.
Determine all the five-digit palindromic integers that are divisible by \(55\).
Note: An integer is divisible by \(11\) exactly when the alternating sum of its digits is divisible by \(11\).
To find the alternating sum, start with the first digit and alternate subtracting and adding the remaining digits from left to right. For example, \(36\,784\) is divisible by \(11\) since \(3-6+7-8+4=0\), and \(0\) is divisible by \(11\). However \(74\,253\) is not divisible by \(11\) since \(7-4+2-5+3=3\), and \(3\) is not divisible by \(11\).
We are looking for all the five-digit integers of the form \(abcba\), that are divisible by \(55\). For an integer to be divisible by \(55\), it must be divisible by both \(11\) and \(5\).
To be divisible by \(5\), an integer must end in \(0\) or \(5\). If a palindrome ends in \(0\), it must also begin with \(0\). However, the integer \(0bcb0\) is not a five-digit integer since the leading digit is \(0\). Therefore, the palindromes cannot end with a \(0\) and hence must start and end with a \(5\). It follows that any five-digit palindrome divisible by \(5\) must be of the form \(5bcb5\).
For an integer to be divisible by \(11\), the alternating sum of its digits must be divisible by \(11\). Therefore, \(5-b+c-b+5=10-2b+c\) must divisible by \(11\). We proceed by looking through the possible values of \(b\).
If \(b=0\), then \(10-2(0)+c=10+c\) must be divisible by \(11\). The only solution is \(c=1\). Thus \(50\,105\) is one of the integers.
If \(b=1\), then \(10-2(1)+c=8+c\) must be divisible by \(11\). The only solution is \(c=3\). Thus \(51\,315\) is one of the integers.
If \(b=2\), then \(10-2(2)+c=6+c\) must be divisible by \(11\). The only solution is \(c=5\). Thus \(52\,525\) is one of the integers.
If \(b=3\), then \(10-2(3)+c=4+c\) must be divisible by \(11\). The only solution is \(c=7\). Thus \(53\,735\) is one of the integers.
If \(b=4\), then \(10-2(4)+c=2+c\) must be divisible by \(11\). The only solution is \(c=9\). Thus \(54\,945\) is one of the integers.
If \(b=5\), then \(10-2(5)+c=c\) must be divisible by \(11\). The only solution is \(c=0\). Thus \(55\,055\) is one of the integers.
If \(b=6\), then \(10-2(6)+c=c-2\) must be divisible by \(11\). The only solution is \(c=2\). Thus \(56\,265\) is one of the integers.
If \(b=7\), then \(10-2(7)+c=c-4\) must be divisible by \(11\). The only solution is \(c=4\). Thus \(57\,475\) is one of the integers.
If \(b=8\), then \(10-2(8)+c=c-6\) must be divisible by \(11\). The only solution is \(c=6\). Thus \(58\,685\) is one of the integers.
If \(b=9\), then \(10-2(9)+c=c-8\) must be divisible by \(11\). The only solution is \(c=8\). Thus \(59\,895\) is one of the integers.
Therefore, there are \(10\) five-digit palindromic integers that are divisible by \(55\). They are \(50\,105\), \(51\,315\), \(52\,525\), \(53\,735\), \(54\,945\), \(55\,055\), \(56\,265\), \(57\,475\), \(58\,685\), and \(59\,895\).