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.

]]>A -> 742 | 742B742 | 783A162

B -> 5 | 162B783 | 5A5

So the first few are:

742

7425742

783742162

74257425742

7421625783742

7837425742162

742574257425742

783783742162162

74216257425783742

74257837421625742

78374257425742162

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. ðŸ™‚

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