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How to calculate Remainder when a Number $73$^$382$ is divided by $100$?

MyApproach

I dont have approach for these type of modal questions.Can Anyone guide me how to approach the problem?

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I got a slightly different answer than "lab bhattacharjee". Which one of us made a mistake?

Given that $73^{20}\equiv 1\pmod{100}$ and $73^{382}=73^{20\cdot 19+2}$ so

$$ \begin{align*} 73^{382}\pmod{100}&\equiv [73^{20}\pmod{100}]^{19}\cdot [73^2\pmod{100}]\\ &\equiv 73^2\pmod{100}\\ &\equiv 29\pmod{100}. \end{align*} $$

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    $\begingroup$ It is congruent to $29$. $\endgroup$ – André Nicolas Nov 3 '15 at 5:46
  • $\begingroup$ @Laars helenius Are you available in chat.I got similar question in which I applied Your Concept.But I am not getting correct Ans. $\endgroup$ – Jack Nov 3 '15 at 9:10
  • $\begingroup$ Hi jack! I am sorry that I had went to bed after my answer and missed your message about a chat. I am available now if you are still interested. $\endgroup$ – Laars Helenius Nov 3 '15 at 12:32
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We use Euler's Theorem. The calculation may not be suited to your needs.

By Euler's Theorem, $73^{20}\equiv 1\pmod{25}$.

Also, $73^2\equiv 1\pmod{4}$, so $73^{20}\equiv 1\pmod{100}$.

It follows that $73^{380}\equiv 1\pmod{100}$, and therefore $73^{382}\equiv 73^2\equiv 29\pmod{100}$.

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As $73^2=5329,$

$73^{382}=(5330-1)^{191}=(-1+5330)^{191}\equiv(-1)^{191}+\binom{191}1(-1)^{191-1}\cdot5330^1\pmod{100}$

As $191\cdot533\equiv3\pmod{10},$

$$191\cdot5330\equiv30\pmod{100}$$

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    $\begingroup$ Did you lose the $(-1)^{191}$ in the middle of your second line? It would make your answer agree with Andre and Laars. $\endgroup$ – alex.jordan Nov 3 '15 at 5:47
  • $\begingroup$ @alex.jordan, $$(-1)^{191}+\binom{191}1(-1)^{191-1}\cdot5330^1=-1+191\cdot5330=-1+30=?$$ $\endgroup$ – lab bhattacharjee Nov 3 '15 at 5:50
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    $\begingroup$ Your display makes it read as if $30$ is your final stated answer. $\endgroup$ – alex.jordan Nov 3 '15 at 5:53
  • $\begingroup$ @alex.jordan, I did want to leave something for the OP $\endgroup$ – lab bhattacharjee Nov 3 '15 at 5:54
  • $\begingroup$ That's fine, but using display mode math at the end has the effect of making it sound like you are saying $30$ is the answer. If that last part were left in inline math mode, your intent would be more clear. $\endgroup$ – alex.jordan Nov 3 '15 at 5:56

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