Finding 5-Digit Numbers w/ Neighbouring Digits Differing by 3

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Discussion Overview

The discussion revolves around finding 5-digit numbers where each pair of neighboring digits differs by 3. Participants explore methods for determining the total count of such numbers, considering both trial and error and potential formulas.

Discussion Character

  • Exploratory, Mathematical reasoning, Homework-related

Main Points Raised

  • One participant questions whether a formula is needed or if trial and error is sufficient to find the numbers.
  • Another participant suggests that if the first digit is 1, the second digit must be determined based on the difference of 3.
  • A participant provides an example of a valid number, 14741, but expresses concern about the number of possibilities.
  • There is a suggestion that a recurrence relation could be written, although one participant doubts its usefulness and encourages a straightforward approach instead.
  • A later reply mentions that this problem was part of a mathematics competition and hints that the answer is around 40.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to solve the problem, with some favoring trial and error while others suggest more structured methods like recurrence relations.

Contextual Notes

There are unresolved assumptions regarding the starting digit and the implications of neighboring digit differences, as well as the potential complexity of counting combinations.

thagamizer
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how many 5 digit numbers are there in which every two neighbouring digits differ by 3?

can you please tell me if i have to do this all by trial and error or is there some sort of formula i need to make to do this
thanks
 
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Trial and error? You mean you're going to guess a number and see if it satisfies the condition?

What if I told you the first digit were a 1? What is the second digit?
 
14741
but there's too many possibilites
 
what i meant is do i need to make a formula to work out the number of combinations
 
There are not too many possibilities. And that isn't the only one that starts with 1. You could write down a recurrence relation, if you wished, but I doubt that will help - just do it, it isn't very hard, and won't take you very long.
 
This was in the UNSW Maths comp. Hint: it is around 40...
 

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