Find last $3$ digits in expansion of $9^{30}$.

Question

My approach:
$9^{30} = (10-1)^{30}$. Now expanding $(10-1)^{30}$ by binomial theorem we get
${^{30}}C_0(10^{30}) - {^{30}}C_1(10^{29}) + \cdots - {^{30}}C_{27}(10^3) + {^{30}}C_{28}(10^2) - {^{30}}C_{29}(10) + {^{30}}C_{30}$
$10^3[ {^{30}}C_0(10^{27}) - {^{30}}C_1(10^{26}) + \cdots - {^{30}}C_{27} ] + {^{30}}C_{28}(10^2) - {^{30}}C_{29}(10) + {^{30}}C_{30}$
$10^3(k) + {^{30}}C_{28}(10^2) - {^{30}}C_{29}(10) + {^{30}}C_{30}$
$10^3(k) +43201$
Now if we are able to prove that $k$ is a positive integer then surely the last three digit will be $201$ , but I am unable to prove that $k$ is a positive integer . please help me.

Answer 1

Your assertion that all but three terms of the expansion of $(10-1)^{30}$ are multiples of $10^3$ (i.e. that $k$ is an integer) is correct

Indeed you have said $k = {}^{30}C_0(10^{27}) - {}^{30}C_1(10^{26}) + \ldots - {}^{30}C_{27}$ and this is clearly an integer since it is the product and sum of integers

To show it is positive, as Lord Shark the Unknown has commented, you need to show $9^{30} \gt 43201$. Since $9^5=59049\gt 43201$, you can be sure $9^{30}$ is bigger

Answer 2

Do you really require $k$ to be positive?
You would require that if $|k|$ was large enough that $10^3k + 43201$ was negative.

However, note that $9^{30}$ is strictly positive to start with. Thus, even if $k$ is negative, $0 < 9^{30} = 10^3k + 43201$ will still have the last three digits are $201$ and thus, you are done.

Answer 3

Yes, you can say that $k$ is an integer and it doesn't matter whether it's positive or negative, because the number you end with when discarding the multiple of $1000$ is positive.

Let's rewrite the expansion in a more standard way: $$ (10-1)^{30}=\sum_{n=0}^{30}\binom{30}{n}(-1)^n 10^{n}=\sum_{n=0}^2 \binom{30}{n}(-1)^n 10^n+1000\sum_{n=3}^{30}\binom{30}{n}(-1)^n 10^{n-3} $$ The second summation is a multiple of $1000$, so we can discard it. Then we get $$ 9^{30}=(10-1)^{30}\equiv 1-\binom{30}{1}10+\binom{30}{2}100\pmod{1000} $$ It just take a minute to show this is $43201\equiv201\pmod{1000}$.

OK, what if we do the same with the last two digits? We have, with a similar method, $$ 9^{30}=(10-1)^{30}\equiv 1-\binom{30}{1}10\pmod{100} $$ which yields $9^{30}\equiv-299\pmod{100}$. Where's the problem? $-299\equiv 1\pmod{100}$, so the last two digits are $01$.