A number is called a palindrome if it reads the same backward as well as forward. For example 285582

Q: Wait, why can't $a$ be $0$? If $a=0$, the number is still a palindrome, right?

That is a common trap! While $05cc50$ is a palindrome, it is not a six-digit number. A six-digit number must have a non-zero leading digit. If $a=0$, the number effectively becomes a five-digit number (or less), which violates the problem statement. Always check your leading digit constraints!

Q: Why does the alternating sum for $11$ always result in $0$?

It is the beauty of the palindrome structure! Because the digits are mirrored, the sum of the odd-positioned digits $(a+c+b)$ is identical to the sum of the even-positioned digits $(b+c+a)$. When you subtract them, they cancel out perfectly to $0$. Since $0$ is divisible by $11$, the condition is satisfied for any choice of $b$ and $c$.

Q: Are there any restrictions on $b$ and $c$?

None at all! Since the divisibility by $11$ is automatically satisfied for any digit, and the divisibility by $5$ is satisfied by fixing $a=5$, the digits $b$ and $c$ are free to take any value from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. This gives us $10$ choices for $b$ and $10$ choices for $c$.

Question

A number is called a palindrome if it reads the same backward as well as forward. For example 285582 is a six digit palindrome. The number of six digit palindromes, which are divisible by 55, is .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

A six-digit palindrome is a number that reads the same forward and backward. We can represent this number using the digit structure $abccba$, where $a$, $b$, and $c$ are the digits.

In this representation, $a$ is the first and sixth digit, $b$ is the second and fifth digit, and $c$ is the third and fourth digit. Since it is a six-digit number, the leading digit $a$ must satisfy the constraint $a \in \{1, 2, \dots, 9\}$.

### The Divisibility Hurdle

The problem requires the number to be divisible by $55$. Since $55 = 5 	imes 11$ and $\gcd(5, 11) = 1$, the number must be simultaneously divisible by $5$ and $11$.

For a number to be divisible by $5$, its last digit must be $0$ or $5$. In our structure $abccba$, the last digit is $a$.

Since $a$ cannot be $0$ (as it is the leading digit of a six-digit number), we must have $a = 5$. Our number now takes the specific form $5bccb5$.

### The Magic of Eleven

A number is divisible by $11$ if the alternating sum of its digits is a multiple of $11$. For the number $5bccb5$, the digits at odd positions are $5, c, b$ and the digits at even positions are $b, c, 5$.

The alternating sum is calculated as follows:

$$S = (5 + c + b) - (b + c + 5)$$

Upon simplification, we observe:

$$S = 5 + c + b - b - c - 5 = 0$$

Since $0$ is a multiple of $11$, the condition for divisibility by $11$ is satisfied for any choice of $b$ and $c$. The divisibility by $11$ imposes no further restrictions on the digits.

### The Final Count

We are left with two independent variables, $b$ and $c$. Each variable can take any integer value from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

There are $10$ possible choices for $b$ and $10$ possible choices for $c$. The total number of such palindromes is the product of these independent choices:

$$N = 10 	imes 10 = 100$$

There are exactly **100** six-digit palindromes divisible by $55$.

Animated Solution for Mathematics - Permutations and Combinations: A number is called a palindrome if it reads the same backward as well as forward. For example 285582 is a six digit palindrome. The number of six digit palindromes, which are divisible by 55, is .

Enter Numerical Value:

Visualized Solution

Defining the Palindrome Structure

Divisibility by $55$

Applying Divisibility by $5$

Constraint: $6$ -Digit Number

Updating the Number Structure

Divisibility by $11$ Rule

Calculating the Alternating Sum

The Result is Always Zero

Counting Choices for $b$

Counting Choices for $c$

Final Calculation

The Sigma Insight: Fundamental Principle of Counting

Analyzing the Setup

The Divisibility Hurdle

The Magic of Eleven

The Final Count

Similar Questions

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