Continue the Sequence: 742, …

This is the sequence of numbers n such that 3 times the reversal of n plus 1 is the number itself. In other words, n = 3*reversal(n)+1. For example, 742 = 3*247+1. In fact, 742 is the smallest number with this property. How does this sequence continue, and why?

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3 Comments

  1. Joseph:

    I figured out that 783742162 is another number in the sequence. So is 783783742162162. In fact, you can keep tacking 783’s at the start and 162’s at the end, and thus get infinitely many such numbers.

    Are those (with 742) all the numbers in the sequence? I don’t know, but I’m guessing yes, since otherwise you probably wouldn’t have asked this question. 🙂

  2. Piotr:

    It seems that the answer is all numbers A that can be generated (recursively) by the following productions:

    A -> 742 | 742B742 | 783A162
    B -> 5 | 162B783 | 5A5

    So the first few are:

    742
    7425742
    783742162
    74257425742
    7421625783742
    7837425742162
    742574257425742
    783783742162162
    74216257425783742
    74257837421625742
    78374257425742162

  3. Uniteasy:

    The number in the largest place must be 7 and the number in the ones place must be 2.

    The number in the ones place after product with 3
    3 x 1 -> 3
    3 x 2 -> 6
    3 x 3 -> 9
    3 x 4 -> 2
    3 x 5 -> 5
    3 x 6 -> 8
    3 x 7 -> 1
    3 x 8 -> 4
    3 x 9 -> 7

    Since n-1 = 3*reversal(n)

    If the number in the ones place of n is 2, the number in the ones place of n-1 is 1, which is the number in the ones place of 7 multiplying by 3. Then 7 is the number in the largest place of n.

    Let n be a 4-digit number, there are the following combinations n and its reversal
    7xx2 2xx7
    6xx3 3xx6
    1xx4 4xx1
    2xx5 5xx2
    5xx6 6xx5
    2xx7 7xx2
    9xx8 8xx9
    6xx9 9xx6

    Since n > reversal(n) and (n-1)/reversal(n) =3, there’s only one possibility for the smallest and largest places combination, that is
    7xx2 2xx7

    Solution for the 4 digit number
    Let a be the number in tens place of n, and b the number of hundreds place.
    n-1 = 3*reversal(n) is converted to the flowing equation
    7000+100a+10b+2-1= 3*(2000+100b+10a+7)
    980+70a = 290b

    No integer pairs are found for a and b. So no solution for 4 digit numbers.

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