# How do I put this series into a formula?

Tags:
1. May 9, 2017

### SamBull

1. The problem statement, all variables and given/known data
Consider now a sample consisting of all possible n-digit integers where n is odd. Use your answers to the first two parts or otherwise to deduce a formula for the number of palindromic numbers within this sample.Your formula should be a function of n only. [6 marks] [Hint: Consider how this number changes as n varies through 3,5,7 etc and study the pattern]

2. Relevant equations
The earlier parts of the exam question asked how many Palindromic numbers there are using 3 digits i.e between 000-999 which I calculated to be 100 palindromic numbers. The other part before this was the same but for 5 digits i.e 00000-99999 which i calculated to be 1000 palindromic numbers.

3. The attempt at a solution
I can see that when there are n = 3 digits then the total number of palindromic numbers is:

N=10^(3-1) (where 3 is n)

and for n = 5 digits

N=10^(5-2)

I don't know how to relate these into a formula for N using n. I've tried a lot of things but nothing seems to work.

P.S This is my first time posting here sorry if I have done anything wrong :<

2. May 9, 2017

### QuantumQuest

The first digit in a palindrome number cannot be zero, so for the 3 - digit palindrome numbers you have a total of 9 possibilities for the first and third digit (as each such number is a palindrome) and 10 possibilities for the second digit i.e. a total of $9 \cdot 10 = 90$ three - digit palindrome numbers. (Following the same logic can you see how many 5 - digit palindrome numbers are there?)

Now, consider the general case of an $n$ - digit palindrome number where $n$ is odd. Using the hint that the problem provides you can you see the pattern i.e. how the symmetry of a palindromic number constrains the possibilities and for how many digits of a such palindrome number? What about the number of digits left?

3. May 13, 2017

### Staff: Mentor

You did just fine. Welcome to Physics Forums!

4. May 17, 2017

### scottdave

So you are not seeing the pattern. You see for n=3: n-1 =2, and n=5, n-2 = 3, so it doesn't seem to be a pattern. If you look at how palindromes are formed: for 3 digits, you have the first digit must equal the last digit, and the middle can be anything. So you have 10 for the first and 10 for the middle = 10x10 = 100.
For 5 digits, you have the first two digits must mirror the last two, so you have 10^2 = 100 for the first 2, and multiply by 10 for the middle. So you should not be subtracting, but adding (powers of 10). But you don't want to add the full n, because you are only using roughly half of the digits in the calculation.
Hopefully this hint points you in the right direction.

There can be debate about whether or not to use leading zeros. In regular numbers such as counting, you would not do it. But there are plenty of examples of where leading zeros are commonplace, such as ZIP codes, odometer readings or combination locks. You should definitely state that you considered leading zeros, or eliminated their possibility in your calculations.