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Find $e \in \{0,1,\ldots,22\}$ such that the product $\prod_{i=6}^{18} i$ is congruent to $e$ modulo $23$.

I have completed a question similar to this before so I thought I knew how to do this but I believe my answer is wrong.

In order to complete this question these are the steps I took:

$$12 \cong -11 \pmod{23}$$

$$13 \cong -10\pmod{23}$$

$$14 \cong -9\pmod{23}$$

$$15 \cong -8\pmod{23}$$

$$16 \cong -7\pmod{23}$$

$$17 \cong -6\pmod{23}$$

$$(6)(7)(8)(9)(10)(11)(-11)(-10)(-9)(-8)(-7)(-6)$$

$$(-)[6\cdot11][7\cdot10][8\cdot9][9\cdot8][10\cdot7][11\cdot6]\pmod{23}$$

$$(-)(66)(70)(72)(72)(70)(66)$$

$$66=20\pmod{23}$$ $$70=1\pmod{23}$$ $$72=3\pmod{23}$$

$$=(-)(20)(1)(3)(3)(1)(20)$$

The next parts are what confuses me because I then use: $[20\cdot3]^2(1)^2 = 60^2$ and $60\cong14\pmod{23}$ so $(-)(14)^2\pmod{23} =(-)(196)\pmod{23}=12$ and then I finally get that $-12\pmod{23}\cong 11\pmod{23}$ as my final answer. I was told that this is wrong and I am unsure why. I am fairly certain the errors begin after I get:${}=(-)(20)(1)(3)(3)(1)(20).$

miracle173
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Pauliene
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  • Your MathJax usage was abominable and I cleaned it up. See my edits. – Michael Hardy Dec 11 '21 at 19:37
  • Where does the minus sign in the outside of your product come from? – Steven Stadnicki Dec 11 '21 at 19:39
  • You should be aware that there are two usage conventions. One of them I suspect originated in about the middle of the 20th century and says $$ \big( a \bmod b\big) =\text{the remainder when $a$ is divided by $b$}. $$ The other goes back to a book by Carl Gauss published in 1801 called Disquisitiones Arithmeticae, and says $$\begin{align} & \text{“} \big( c \equiv d\big) \bmod e\text{”} \ & \text{ means $c$ and $d$ both leave the same remainder when divided by $e$.} \end{align} $$Notice that the binary relation symbol in that second expression is $\text{“} {\equiv} \text{”}.\qquad$ – Michael Hardy Dec 11 '21 at 19:53
  • A less complicate title would be "calculate $\prod_{i=6}^{18} i \pmod{23}$ – miracle173 Dec 11 '21 at 20:31
  • Hi @MichaelHardy, I tried to make it as clear as possible but I am not as proficient in MathJax as I would like to be. I apologize – Pauliene Dec 11 '21 at 20:38

1 Answers1

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Recall for a prime $p$, $(p-1)! \equiv_{p} -1 $. It follows $$2 \cdot 3 \cdot 4 \cdot 5\cdot \left( \prod_{i=6}^{18} i\right)\cdot (-4) \cdot (-3) \cdot (-2) \cdot (-1)\equiv_{23} (-1)$$ $$\Rightarrow 5 \cdot \left(\prod_{i=6}^{18} i \right) \equiv_{23} -1 \quad \Rightarrow \prod_{i=6}^{18} i \equiv_{23} 9 $$

MyMolecules
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  • Hi, thank you for your response, it has helped me a lot. Do you mind explaining how you got your final answer from the step before it? Thank you. – Pauliene Dec 11 '21 at 20:27
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    @DanaLyn I noticed that $5 \cdot 9 = 45 \equiv_{23} -1$. Essentially I looked for multiples of $23$ with units digit $6$ or $1$ and $46$ worked immediately. – MyMolecules Dec 11 '21 at 20:30
  • Thank you so much! – Pauliene Dec 11 '21 at 20:37