This question already has an answer here:
Is the number 3776543210123456773, in base 8, divisible by 7?
I started by expanding the number:
3 * $8^0$ + 7 * $8^1$ + 7 * $8^2$ + ...
3 * $(7 + 1)^0$ + 7 * $(7 + 1)^1$ + 7 * $(7 + 1)^2$ + ...
Then, by using Newton's Binomial Theorem, I reasoned that for each binomial, only the first term wouldn't be divisible by 7 since it would be equal to 1. This led me to the conclusion that each digit of the initial number would be multiplied by one once. Therefore, if the sum of all digits was divisible by 7, the number would also be divisible by 7.
However, I converted this number to base 10 and the remainder of the division by 7 was equal to 2, instead of 0, which was the result I found using the reasoning above.
I can't seem to find where I made a mistake. :c