Could any one tell me how to find the last two digit of $7^{81}$? I have succeeded in finding the last digit only which is $7$.
Any group theoretic approach or any other approach is welcome.
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http://math.stackexchange.com/questions/362012/find-the-last-two-digits-of-781 – lab bhattacharjee Apr 18 '13 at 06:08
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Recall Euler's theorem. If $\gcd(a,n) = 1$, we have $$a^{\phi(n)} \equiv 1 \pmod{n}$$ For your problem, take $a = 7$ and $n=100$.
We have $\phi(100) = 40$. We then get that $$7^{40} = 7^{\phi(100)} \equiv 1 \pmod{100}$$ This gives us$$7^{81} \equiv 7^{80} \cdot 7 \pmod{100} \equiv \left(7^{40} \right)^2 \cdot 7 \pmod{100} \equiv 7 \pmod{100}$$
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2@miosaki Yes. The last two digits are $07$. If a number leaves a remainder $7$ on divided by $100$, the last two digits are $07$. – Apr 18 '13 at 06:06
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An alternative method of determining the value that does not rely on calculating $\phi(100)$ is to perform the calculation in modular arithmetic using efficient exponentiation.
$81=64+16+1$, so here's the process (all lines are mod 100):
$$7 \equiv 7\\7^2 \equiv 49\\7^4 \equiv 49^2 \equiv 1\\7^5 \equiv 1\cdot 7 \equiv 7\\7^{10} \equiv 7^2 \equiv 49\\7^{20} \equiv 49^2 \equiv 1\\7^{40} \equiv 1^2 \equiv 1\\7^{80} \equiv 1^2 \equiv 1\\7^{81} \equiv 1\cdot 7 \equiv 7$$
This approach works for essentially any set of base, exponent, and modulo number.
Glen O
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