1. Form of a six-digit palindrome

A six-digit palindrome has the form $\overline{abc\,cba}$ where:

• $a \in \{1,2,\dots,9\}$ (since it is a six-digit number),
• $b,c \in \{0,1,2,\dots,9\}$.

So the number is $N = 100000a + 10000b + 1000c + 100c + 10b + a.$

2. Divisibility by 55

Since $55 = 5 \times 11,$ $N$ must be divisible by both $5$ and $11$.

3. Condition for divisibility by 5

A number is divisible by $5$ if its last digit is $0$ or $5$.

In the palindrome $\overline{abccba}$, the last digit is $a$. Also $a$ is the first digit, so $a \neq 0$. Hence the only possibility is $a=5.$

4. Condition for divisibility by 11

For divisibility by $11$, the difference between the sum of digits in alternating places must be a multiple of $11$.

For $\overline{abccba}$:

• Sum of digits in odd places $= a + c + b$,
• Sum of digits in even places $= b + c + a$.

Their difference is $(a+c+b) - (b+c+a) = 0,$ which is always divisible by $11$.

So every even-digit palindrome is divisible by $11$. Thus once $a=5$, divisibility by $55$ is automatically satisfied.

5. Count the palindromes

Now:

• $a=5$ fixed,
• $b$ can be any digit from $0$ to $9$ → $10$ choices,
• $c$ can be any digit from $0$ to $9$ → $10$ choices.

Therefore total number of such palindromes is $10 \times 10 = 100.$

6. Final Answer

The number of six-digit palindromes divisible by $55$ is $\boxed{100}.$