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What is the shortest way to compute the last 3 digits of $17^{256}$ ?

My solution: \begin{align} 17^{256} &=289^{128} \\ &=(290 - 1)^{128}\\ &=\binom{128}{0}290^{128} - ... +\binom{128}{126}290^2 - \binom{128}{127}290 + \binom{128}{128} \end{align}

Computed the last 3 terms whose last 3 digits gave the last 3 digits of $17^{256}$.

Is there any shorter method to do this(which requires much less computation) ?

user2369284
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4 Answers4

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Your way seems to be the fastest for me

$$\binom{128}{128}=1\equiv1\mod{1000}$$

$$\binom{128}{127}=128\equiv28\pmod{100}$$ $$\implies \binom{128}{127}290\equiv 28\cdot290\pmod{1000}\equiv120$$

$$\binom{128}{126}=\frac{128\cdot127}2\equiv8\pmod{10}$$ $$\implies \binom{128}{126}290^2\equiv 290^2\cdot8\pmod{1000}\equiv800 $$

Using $a\equiv b\pmod m\implies a\cdot c\equiv b\cdot c\pmod {c\cdot m}$ where $a,b,c,m$ are integers

lab bhattacharjee
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Use Euler's Totient Theorem and the Chinese Remainder Theorem.

We have that $ 17 \equiv 1 \mod 8 $ and hence $ 17^{256} \equiv 1 \mod 8 $ as well.

Because $ 17 $ is coprime to $ 125 $, we know that $ 17^{100} \equiv 1 \mod 125 $.

We are left to calculate $ 17^{56} \mod 125 $, which can be quickly be done by hand via the method of doubling: $$\begin{align*} 17^{2} &\equiv 39 &\mod 125 \\ 17^4 &\equiv 21 &\mod 125\\ 17^8 &\equiv 66 &\mod 125\\ 17^{16} &\equiv 106 &\mod 125\\ 17^{32} &\equiv 111 &\mod 125 \end{align*}$$

Hence, $ 17^{32 + 16 + 8} \equiv 111 \cdot 106 \cdot 66 \equiv 56 \mod 125 $.

Using the Chinese Remainder Theorem, we have that the list three digits are $ 681 $.

P L Patodia
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Jon Claus
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  • How have you written $17^{100} \equiv 1 \mod 125$ – user2369284 Dec 27 '13 at 18:53
  • Euler's Totient Theorem says that for coprime $ a, b $, $ a^{\phi(b)} \equiv 1 \mod b $, where $ \phi $ is the totient function. In this case $, a = 17, b = 125, $ and $ \phi(b) = 125 \left(1 - \frac{1}{5}\right) = 100 $. – Jon Claus Dec 27 '13 at 18:55
  • How have you done the doubling ? – user2369284 Dec 27 '13 at 18:57
  • Perhaps I should have called it squaring. I simply meant it is easiest to square the number and calculate the new modulus each time. Then you can just write the exponent of the original number in binary and multiply to get the residue. – Jon Claus Dec 27 '13 at 19:08
  • If going by hand use $106\equiv -19$ and $111\equiv -14$ to reduce arithmetic. Also $-59$ is easier to square than $66$ unless you know two digit squares. – Mark Bennet Jan 23 '15 at 16:57
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I am not sure what "shorter" means, but $\phi(1000) = 400,$ so $17^{400} = 1,$ so $17^{200} = \pm 1,$ so you only need to compute $17^{56}.$ Since you don't know which one of $\pm 1$ to take, you have two possible answers, but to know which one is right, look at the last digit.

Igor Rivin
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HINT:

As $1000=125\cdot8$

$$17\equiv1\pmod8\implies 17^n\equiv1\pmod8$$

Now, $\phi(125)=100\implies 17^{256}\equiv17^{56}\pmod{125}$

Then apply CRT

lab bhattacharjee
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