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Find the last $3$ digits of $3^{352}$ ?

Apart from Carmichael Function, any other way of solving it ?

Jon Garrick
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6 Answers6

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Note that $3^4=81$, so $$\begin{aligned}3^{352} &\equiv (80+1)^{88} \pmod{1000} \\ &\equiv 1+\binom{88}{1} 80 + \binom{88}{2} 6400\pmod{1000} \\&\equiv 1+040+200 \pmod{1000} \end{aligned} $$

Because last two digits of $88\times 8$ is $04$, while last digit of $44\times 87\times 64$ is $2$.

pisco
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Note that we can achieve the result with a simple desk calculator with limited capacity just by dichotomy on the exponent.

$352=11\times2^5$

So we first calculate $3^{11}=81\times 81\times 27=177147\equiv 147\pmod{1000}$

Then we successively square five times the result modulo $1000$.

$147^2=21609\to 609^2=370881\to 881^2=776161\to 161^2=25921\to 921^2=848241$

And the final result is $241$.

It required only $7$ multiplications, compared to the $3$ in the smart method by pisco125. So it is not a big overhead.

zwim
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Here is an elementary method (there may be more clever ones):

Notice

$3^{352} = 3^{(11 \cdot 2^5)} = 177147^{(2^5)} = (t + 147)^{(2^5)}$

where $t$ stands for "thousands"

Now as you are only interested in the last 3 digits, you are not interested in any multiples of $t$. With that understanding, we have five times the quadratic equation:

$(t_1 + 147)^2 = t_2' + 147^2 = t_2' + 21609 = t_2 + 609$

$(t_2 + 609)^2 = t_3' + 609^2 = t_3' + 370881 = t_3 + 881$

$(t_3 + 881)^2 = t_4' + 881^2 = t_4' + 776161 = t_4 + 161$

$(t_4 + 161)^2 = t_5' + 161^2 = t_5' + 25921 = t_5+ 921$

$3^{352} = (t_5 + 921)^2 = t_6' + 921^2 = t_6' + 848241 = t_6 + 241$

So the last three digits are 241.

Andreas
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As you imply, the Carmichael function $\lambda(1000) = \text{lcm}(\,\lambda(5^3), \lambda(2^3)\,) = \text{lcm}(100,2) = 100$ is the least universal exponent.

All$\bmod 1000$, this gives us that $3^{352}\equiv 3^{52}$ and then:
$\begin{align}3^{52} = 9^{26} = (10-1)^{26} & \equiv (-1)^{26} + 26\cdot 10\cdot (-1)^{25} + \overbrace{\frac{26\cdot25}2}^{\text{odd mult of 5}}\cdot 10^2\cdot (-1)^{24}\\ & \equiv 1-260+500 \equiv 241\\ \end{align}$

Joffan
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One has $3^4 \equiv 1$ (mod 8) so $3^{352} \equiv 1$ (mod 8).

Moreover, $3^{50} \equiv -1$ (mod 125) so $3^{352} \equiv (-1)3^{2} = -9$ (mod 125).

So $3^{352} - 241 \equiv 0$ (mod 125) and (mod 8). So $3^{352} \equiv 241$ (mod 1000).

GAVD
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  • There's a typo, it should be $3^{50} \equiv -1\pmod{125}$, as the next formula shows. –  Sep 29 '17 at 08:43
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    can you detail how you came to $3^{50}\equiv -1\pmod{125}$ ? – zwim Sep 29 '17 at 08:53
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    Here is one way that does not involve the theory of quadratic residues. Start with $3^5=243\equiv -7\bmod 125$. Square to obtain $3^{10}\equiv 49=50-1$. Expand $3^{50}\equiv (50-1)^5$ with the binomial theorem and note that only one term has nonzero residue $\bmod 125$. – Oscar Lanzi Sep 30 '17 at 10:24
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Take $3^{352}\equiv x\pmod{10}$.

Which is equivalent to

$\begin{align}({3^{4}})^{88}&\equiv x\pmod{10}\\81^{88}&\equiv x\pmod{10}\\1^{88}&\equiv x\pmod{10}\\\end{align}$

Therefore, $x\equiv1\pmod{10}$. Since the last digit of any $n$ is $n\pmod{10}$, so the last digit of $3^{352}$ is $1$.

DynamoBlaze
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