My approach:
$9^{30} = (101)^{30}$. Now expanding $(101)^{30}$ 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.
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5You have been a member for 3 months and asked 27 questions. It is time to start formatting your questions properly! – almagest Jan 05 '20 at 13:47

3I reckon that $9^{30}$ has just got to be a bit bigger than $43201$. – Angina Seng Jan 05 '20 at 13:48

Sorry for not formatting my question . Actually I have my jee mains exam day after tommorrow . I could not learn how to format the questions due to preparation for the exam . Please help me for just 2 more days – Sameer nilkhan Jan 05 '20 at 13:50

https://math.stackexchange.com/questions/355864/patterntolastthreedigitsofpowerof3 – lab bhattacharjee Jan 05 '20 at 13:58

What do you denote $k$, exactly? – Bernard Jan 05 '20 at 14:26

it is an integer. – Sameer nilkhan Jan 05 '20 at 14:30
3 Answers
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.
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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: $$ (101)^{30}=\sum_{n=0}^{30}\binom{30}{n}(1)^n 10^{n}=\sum_{n=0}^2 \binom{30}{n}(1)^n 10^n+1000\sum_{n=3}^{30}\binom{30}{n}(1)^n 10^{n3} $$ The second summation is a multiple of $1000$, so we can discard it. Then we get $$ 9^{30}=(101)^{30}\equiv 1\binom{30}{1}10+\binom{30}{2}100\pmod{1000} $$ 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, $$ 9^{30}=(101)^{30}\equiv 1\binom{30}{1}10\pmod{100} $$ which yields $9^{30}\equiv299\pmod{100}$. Where's the problem? $299\equiv 1\pmod{100}$, so the last two digits are $01$.
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1It _does_ matter whether $k$ is positive, and the way I read it, that is the single thing that the OP was struggling with. The fact that you just brush that away in a side note makes this answer not good in my opinion. For instance, $799$ is also congruent to $201$ modulo $1000$, but has very different final three digits. – Arthur Jan 05 '20 at 14:05


Well, you didn't even deign to spare that much effort into explaining the one thing the OP was stuck on. If that comment had been in your answer post, I would have no quarrel. In fact, if that comment had been _all_ of your post, it would still have been a decent answer IMO. – Arthur Jan 05 '20 at 14:11

Your assertion that all but three terms of the expansion of $(101)^{30}$ 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
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