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?
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
I agree.
Thanks a bunch!
To determine whether a given number is divisible by 3, we can check if the sum of its digits is divisible by 3. So, in order to answer this question, we need to determine the sum of the 99 consecutive digits from the decimal form of 1/13.
To find the decimal form of 1/13, we can perform a long division:
```
0.076923076923076923...
```
We can see that the decimal repeats in a cycle of 6 digits: 076923. So, the 99 consecutive digits will consist of multiple repetitions of this cycle.
To find the sum of the digits, we can add up the individual digits of each repetition of the cycle. The sum of the digits in the cycle is:
0 + 7 + 6 + 9 + 2 + 3 = 27
Since the sum of the digits in the cycle is 27, which is divisible by 3, we can conclude that the sum of the 99 consecutive digits will also be divisible by 3.
Therefore, the number formed by the 99 consecutive digits from the decimal form of 1/13 will be divisible by 3.