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

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To find 5-digit numbers where each pair of neighboring digits differs by 3, a systematic approach is recommended rather than trial and error. Starting with a specific digit, such as 1, leads to a limited set of valid second digits, which can be expanded to find all combinations. While a recurrence relation could be formulated, it may not simplify the process significantly. The discussion suggests that the total number of such 5-digit numbers is approximately 40. Ultimately, a direct method of generating combinations is deemed efficient for this problem.
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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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