Got his question: (a) Use the binomial theorem to expand (x+1)^99, and show that (x+1)^99 = x^2f(x) + 99x + 1, where f(x) is a function in x.
(b) Using the result in (a), find the remainder when 7^99 is divided by 9.
The answer of (b) is 1. I got 1/9, why am I wrong?
On
$$(x+1)^{99}=\sum_{k=0}^{99}\binom{99}k\,x^k=1+\binom{99}1\,x+\binom{99}2x^2+\sum_{k=2}^{99}\binom{99}k\,x^k=$$
$$1+99x+x^2\left(\binom{99}2+\sum_{k=2}^{99}\binom{99}k\,x^{k-2}\right)$$
This proves (a), and as for (b):
$$7^{99}=(6+1)^{99}\stackrel{(a)}=1+99\cdot6+6^2f(6)=1\pmod 9$$
since $\;99\;$ and $\;6^2\;$ are divisible by $\;9\;$ .
$\frac {10}9 = 1 + \frac 19$
Here the remainder is $1$ and not $\frac 19$
Just a matter of fact - Remainders can never be fractions. They have to be whole numbers.