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The question is:

Find the last two digits of the expansion of $2^{12n}-6^{4n}$ where $n$ is any positive integer.

If we put the value of $n=1$ we would get $2800$. For $n = 2$ the result will be $15097600$ so the last two digits are $00$. Now this approach (using induction) is fairly simple but a bit tedious and time consuming. In my module there is another approach given which proceeds by representing $2^{12n}-6^{4n}$ as $64^{2n}-36^{2n}$ and then as this is divisible by $(64+36)=100$, it says hence the last two digits will be $00$. Now I suppose this is just because anything that is divisible by 10 or 100 would have last digits ending with zeroes. If so then it is not possible to get the last digits of expansions which don't have $10$ or $100$ as a factor. That is, last digits would be anything but zero. Am I correct?

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up vote 2 down vote accepted

Your question is somewhat vague in that you don't say by what means such digits might be inferred. For instance, if you can deduce that a number is divisible by $5$ and that it is odd, then you can deduce that the last digit must be $5$. But if you're asking specifically whether there is any number $n$ that's not divisible by $10$ such that knowing that a number is divisible by $n$ allows you to infer its last digit, the answer is indeed "no".

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What I mean if I can deduce a number is divisible by $10$ then the last digit is $0$ and if a number is divisible by $100$ then the last digits are $00$ right? But this could not be possible for other digits for example say that above expansion is also divisible by $28 (64-36)$ but it does not say anything about the last digits. – Quixotic Jun 20 '11 at 12:21
I don't understand. Are you saying something different from what I wrote in the the last sentence? It sounds like the same thing to me -- if not, please point out where exactly it differs. – joriki Jun 20 '11 at 12:26
Yes,it is the same thing,Thanks. – Quixotic Jun 20 '11 at 12:29

As joriki says, your question is somewhat vague, so I'll give a complementary answer. The last two digits of $7^{4n}$ are guaranteed to be 01, since $7^{4n}=(7^4)^n=(2401)^n=(1+2400)^n$ and expanding the last term give you 1 plus a lot of terms that end in 00. So there are cases where you can get the last digits of expansions even though they don't have 10 or 100 as a factor.

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Yes,I can understand now,for $7^{4n}$ I would have proceed a bit differently,the cyclicity of $7$ and as $4n%4=0$ hence the last digit of $7^{4n}$ is equal to the last digit of $7^{4}$ which is $1$ – Quixotic Jun 20 '11 at 12:27

I am not sure how really specific is your question but I hope you are aware that there is a standard technique to solve problems like :

What are the last two digits of $\langle\hbox{some number defined by a complicated expression}\rangle$.

This technique is called modular arithmetic and consists--in down-to-earth terms--of operating sistematically with the remainders of the numbers involved left by a division by a fixed number $N$ (the modulus).

In the case of the last-two-digits problem, the modulus is $N=100$, just because the last two digits is what is left after cancelling out a suitable multiple of $100$.

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I am aware of the idea and but I never (yet) did using the same yet,however if you like to refer some material where the idea is discussed in detail with few worked out and practice exercise,I would be grateful. – Quixotic Jun 22 '11 at 6:20
Modular arithmetic is discussed in every first (I'm tempted to say zero-th) Algebra course textbook. I'm sure you can find plenty just googling. You can as well start from Wikipedia's page – Andrea Mori Jun 22 '11 at 8:16

As $a^n-b^n$ is divisible by $a-b$ where $a,b,n$ are positive integers

So, $(2^{12n}-6^{4n})$ is divisible by $(2^{12}-6^{4})=(2^6+6^2)(2^6-6^2)=100\cdot(-28)$ hence by $2800$


Clearly, $2^{12n}-6^{4n}$ is divisible by $4$

Now, $2^6=64\equiv -11\pmod{25}\implies 2^{12}\equiv(-11)^2\equiv-4\pmod{25}$

Similarly, $6^2=36\equiv11\pmod{25}\implies 6^4\equiv 11^2\equiv-4\pmod{25}$

So, $2^{6n}-6^{4n}\equiv (-4)^n-(-4)^n\pmod 25\equiv0$

So, $25$ divides $(2^{6n}-6^{4n})$

$\implies $lcm$(25,4)$ divides $(2^{6n}-6^{4n})$

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