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Is there a fast technique that I can use that is similar in each case to calculate the following:

$$(1100)^{1357} \mod{2623} = 1519$$

$$(1819)^{1357} \mod{2623} = 2124$$

$$(0200)^{1357} \mod{2623} = 2227$$

$$(1111)^{1357} \mod{2623} = 1111$$

I used Wolfram Alpha to get to these answers, but I would like to know how to calculate it by hand (with a standard pocket calculator).

Martin Sleziak
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Julia
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4 Answers4

3

I would follow the pseudocode given here:

  • Write your exponent, 1357, in binary: $10101001101_2$.
  • Let $b := x\mod 2623$.
  • Let $r := 1$.
  • Step through the bits from right-to left:
    • If the bit is $1$:
      • Let $r := b \cdot r \mod 2623$.
    • Let $b := b^2 \mod 2623$.

Then $r$ will be your final result. This requires 17 multiplications modulo $2623$. In general it requires (number of 1-bits) + (number of bits) multiplications.

(To compute a "modulo" operation on a pocket calculator, you can either subtract $2623$ repeatedly until you get a result that's less than $2623$, or you can calculate $x - \lfloor x / 2623 \rfloor \cdot 2623$.)

EDIT: You can use the Carmichael theorem to reduce the exponent to 97, as @wythagoras explains. Cool!

Community
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Lynn
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2

Use the Carmichael theorem. This theorem states that $$a^{\lambda(n)} \equiv 1 \mod n$$

if $\gcd(a,2623)=1$. In this case we have $\lambda(2623)=\mathrm{lcm}(42,60)=420$.

Therefore, if $\gcd(a,n)=1$, then $$a^{1357} \equiv a^{1357-1260} = a^{97} \mod 2623$$

This is more friendly to compute with the method @Mauris describes.

wythagoras
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0

Factorize $2623=43\times 61$. Remember the Euler theorem, which tells you that

$$a^{\phi(n)}=1\mod n$$ if $n$ and $a$ are coprime. For prime $n$, you have $\phi(n)=n-1$.

You can also use the CRT

orion
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    How do you combine the results? $1111^{1357} \equiv 36 \pmod{43}$ and $1111^{1357} \equiv 13 \pmod{61}$ -- how do you get from those two numbers to $1111$? – Lynn Jul 20 '15 at 13:06
  • The solution is in the form that satisfies $x=36+43n=13+61m$ for integers $n$, $m$. This is a linear diophantine equation that can be solved with standard algorithms. – orion Jul 20 '15 at 13:08
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Combine Little Fermat and Cinese remainder theorem: as $2623=43\cdot61$, we have: $$\mathbf Z/2623\mathbf Z\simeq \mathbf Z/43\mathbf Z\times \mathbf Z/61\mathbf Z.$$ To compute, say, $1100^{1357}\bmod 2623$, you first compute $1100\bmod 43=25$ and $1100\bmod61=2$. Then use Little Fermat for each of the congruences: $$\begin{cases} 1100^{1357}\equiv 25^{1357\bmod42}=25^{13}\equiv 14\mod 43\\ 1100^{1357}\equiv 2^{1357\bmod60}=2^{37}\equiv 55\equiv-6\mod 61 \end{cases}$$ (the powers can be computed with fast exponentiation, by successive squarings).

Now we have to go back to $\mathbf Z/2623\mathbf Z$. From the Bézout relation: $\;12\cdot61-17\cdot43=1\;$ (obtained from the extended euclidean algorithm, we obtain: $$1100^{1357}\equiv 12\cdot14\cdot61-17\cdot55\cdot43\equiv1819\mod 2623.$$

Bernard
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