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This question covered large exponents on the $b$ side. What about the $m$ side?

Given:

$$a^b \pmod m$$

where $m$ is a large compound number.

For example:

Given

$$5^{2003} \pmod {7} \equiv 3$$ $$5^{2003} \pmod {11} \equiv 4$$ $$5^{2003} \pmod {13} \equiv 8$$

how can one quickly deduce:

$$5^{2003} mod (7*11*13)$$

ChocolateOverflow
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1 Answers1

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Let the number be $x$.

Then we get from the Chinese Remainder Theorem: $$5^{2003}\pmod{7\cdot 11\cdot 13}\equiv x\iff\begin{cases}x\pmod 7\equiv 3\\x\pmod{11}\equiv 4\\ x\pmod{13}\equiv 8\end{cases}$$

The following is one method to apply the Chinese Remainder Theorem.

From the 3rd equation: $$x=13k+8\tag 4$$

Combine with the 1st equation: $$13k+8\equiv -k+1\equiv 3\pmod 7\implies k=7l-2$$ Substitute in (4): $$x=13(7l-2)+8=13\cdot7l-18\tag 5$$ Combine with 2nd equation: $$13\cdot7l-18\equiv 3l+4\equiv 4\pmod{11} \implies l=11m$$ Substitute in (5): $$x=13\cdot7\cdot 11m -18 \equiv -18\pmod{7\cdot 11\cdot 13}$$

Klaas van Aarsen
  • 5,908
  • Where exactly does $5^{2003}$ come in here? – ChocolateOverflow Jan 15 '20 at 04:03
  • @JohnZhau, the 3 given equations are derived from it. E.g. $5^{2003}\pmod 7\equiv (5^{333})^6\cdot 5^5\equiv 1\cdot (-2)^5\equiv 3$. Since you already have those equations, we don't see it anymore. – Klaas van Aarsen Jan 15 '20 at 07:11
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    The chinese remainder theorem says there is only one such value $x$ that is congruent to those values and mods. So as $5^{2003}$ is congruent to those values, it must be that $5^{2003} \equiv x \pmod {71113}$. So solve for $x$ and that will solve for $5^{2003}$. – fleablood Jan 15 '20 at 07:38
  • tl;dr.... Let $x = 5^{2003}$. Solve for $x$. $x\equiv -18\pmod{71113}$ therefore..... $5^{2003} \equiv -18\pmod{71113}$ – fleablood Jan 15 '20 at 07:39