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What are the last two digits of $3^{100}11^{50}+7^{518}$?

I know that to find the last two digits implies that you are essentially finding the remainder when divided by $100$.

Can I simplify $3^{100}$ to $3$ mod $1$ by dividing both the exponent and $100$ (which are the same in this case) by $100$?

Also, how would I simplify $7^{518}$ if that were the case?

Any help is appreciated; thanks!

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  • $\begingroup$ you use the tag modular-arithmetic, do you know Euler's Theorem? $\endgroup$ – lulu Mar 4 '19 at 18:20
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    $\begingroup$ "Can I simplify $3^{100}\pmod{100}$ to $3\pmod{1}$ by dividing both the exponent and $100$ by $100$?" No. $a^{b}\pmod{n}$ is not the same thing as $a^{b/n}\pmod{1}$. $\endgroup$ – JMoravitz Mar 4 '19 at 18:22
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    $\begingroup$ Also: helps to remark that $3^{100}=9^{50}$. $\endgroup$ – lulu Mar 4 '19 at 18:22
  • $\begingroup$ Hint: consider $$3^{100}\cdot 11^{50}+7^{518}$$ $$\mod 100$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 4 '19 at 18:25
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    $\begingroup$ @Dr.SonnhardGraubner That is hardly a hint since the OP already noted this in the post. Also, you went and changed the exponent of the $7$, most likely accidentally. $\endgroup$ – JMoravitz Mar 4 '19 at 18:26
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As you suggested, we want some $0\leq n \leq 99$ such that $$ n \equiv 3^{100} \cdot 11^{50} + 7^{518} \pmod{100}. $$ As noted by lulu in the comments, we first observe that $3^{100} = 9^{50}$. Hence, we have \begin{align} n &\equiv 3^{100} \cdot 11^{50} + 7^{518} \pmod{100} \\ &= 9^{50} \cdot 11^{50} + 7^{518} \\ &= 99^{50} + 7^{518} \end{align}

Note that $99^2 \equiv 1 \pmod{100}$ and $7^4 \equiv 1 \pmod{100}$, so \begin{align} n &\equiv 99^{50} + 7^{518} \pmod{100}\\ &= (99^2)^{25} + (7^4)^{129} \cdot 7^2 \\ &\equiv 1^{25} + 1^{129} \cdot 7^2 \pmod{100} \\& = 1 + 49 =50. \end{align}

Therefore, the last two digits of $3^{100} \cdot 11^{50} + 7^{518}$ are $5$ and $0$.

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$$ 3^{100}11^{50} = 9^{50}11^{50} = 99^{50}$$

$$ 99^{50} (mod 100) = (-1)^{50} (mod 100) = ((-1)^2)^{25} = 1^{25} = 1$$

$$ 7^4 (mod 100) = 2401 (mod 100) = 1 $$

$$ 7^{518} (mod 100)= 7^2(7^{4})^{129} (mod 100) = 7^2 = 49 $$

$$ 1 + 49 = 50 $$

Then, the result is: $$50$$

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  • $\begingroup$ $\large \bmod 100\!:\,\ 7^{\Large 4} \equiv (50\!-\!1)^{\Large 2} \equiv \color{#c00}{50^2}\!-\color{#0a0}{2(50)}+1\,\equiv\, \color{#c00}0-\color{#0a0}0+1\,\equiv\, 1\ \ $ $\endgroup$ – Bill Dubuque Mar 4 '19 at 19:39
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As $\phi(100)=100\cdot\frac{1}{2}\cdot\frac{4}{5}=40$, we know (by Euler’s Theorem) that $$7^{40}\equiv1\pmod{100}$$ Hence, $$7^{518}\cdot7^2=7^{520}=(7^{40})^{13}\equiv1\pmod{100}$$ And as the inverse of $7\pmod{100}$ is $43$, $$7^{518}\equiv1\cdot43^2=1849\equiv49\pmod{100}$$

Now we can calculate $$3^{100}\cdot11^{50}+7^{518}$$ $$\equiv(9\cdot11)^{50}+49\pmod{100}$$ $$\equiv(-1)^{50}+49\pmod{100}$$ $$\equiv1+49\pmod{100}$$ $$\equiv50\pmod{100}$$

So the last two digits of $3^{100}\cdot11^{50}+7^{518}$ are $50$.

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