**Quote:**

The repeating terms i can understand

We do not have the repeating terms themselves, but we have repeating two last digits.

2,

7,

22,

67, 2

02, 6

07, 18

22, 54

67.

**Quote:**

but how did you figure out that the tens and units of the number is 2?

We know that the terms of the sequence have a pattern:

...02, ...07, ...22, ...67, ...02, ...07, ...22, ...67, ...02, ...07, ...22, ...67.

So the four endings (02, 07, 22, 67) repeat. Therefore we know that the 4th, 8th, 12th, ... (4

*n*) -th terms end with 67; the 3rd, 7th, 11th, ... (4

*n* + 3)-th terms end with 22, etc.

35 = 32 + 3 = 4 × 8 + 3

So the 32nd term ends with 67; 33rd term ends with 02; 34th term ends with 07 and the 35th term ends with 22.

-----------------------

**The strict mathematical proof of why this pattern will hold**:

The last two digits of the resulting number in case of multiplication by 3 and addition of 1 depend solely on the last two digits of the original number.

Suppose

*R* =

*ab*....

*cxyz*, where each letter represents a digit, and there could be any number of any digits in between

*b* and

*c*. We can rewrite

*R* as:

*R* =

*ab*....

*cxyz* =

*ab*....

*cx*00 +

*yz*We can see that digits

*ab*....

*cx* do not affect last two digits if

*R* is multiplied by an an integer or if to add a positive number.