How do I put this series into a formula?

  • Thread starter Thread starter SamBull
  • Start date Start date
  • Tags Tags
    Formula Series
Click For Summary

Homework Help Overview

The discussion revolves around finding a formula for the number of palindromic numbers within a sample of all possible n-digit integers, specifically when n is odd. Participants are encouraged to deduce a function of n based on observed patterns from smaller cases, such as 3-digit and 5-digit palindromes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations for 3-digit and 5-digit palindromic numbers, noting the patterns in the number of possibilities for each digit. There is an exploration of how to generalize these findings into a formula based on n.

Discussion Status

Some participants have provided hints and guidance regarding the relationship between the digits in palindromic numbers and how to account for leading digits. There is an ongoing exploration of the symmetry in palindromes and how it affects the total count, with no explicit consensus reached yet.

Contextual Notes

Participants note the importance of considering leading zeros in their calculations, as this could affect the total number of palindromic numbers. There is also mention of homework constraints that may limit the approach to the problem.

SamBull
Messages
1
Reaction score
0

Homework Statement


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]

Homework 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.

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 :<
 
Physics news on Phys.org
SamBull said:

Homework Statement


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]

Homework 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.

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 :<

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?
 
SamBull said:
P.S This is my first time posting here sorry if I have done anything wrong :<
You did just fine. Welcome to Physics Forums!
 
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.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
55
Views
7K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 18 ·
Replies
18
Views
3K