## Continue the Sequence: 742, …

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.