From the decimal form of 1/13 we pick 99 consecutive digits and treat it as a 99-digit number. Can this number be divisible by 3?

Answer 1

1/13 is a repeating decimal,
.076923 076923 07... the length of the repetition is 6 and sum of each loop is 27

so as long as we add up exact number of loops, we would have as sum a multiple of 27, clearly divisible by 3

but 99/6 is not a whole number, namely 16.5
that means we would add up 16 complete loops plus the next 3 digits of wherever we started.

Now, no matter where we started, the next 3 digits are not a multiple of 3.
e.g.
7+6+9 = 22
6+9+2 = 17
9+2+3 = 14
2+3+0 = 5
3+0+7 = 10
0+7+6 = 13

So, no, the sum of any consecutive 99 digits is not divisible by 3

Answer 2

I agree.

Answer 3

Thanks a bunch!