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In the prime factorization of $109!$, what is the exponent of $3$? (Reminder: The number $n!$ is the product of the integers from 1 to $n$. For example, $5!=\(5\cdot 4\cdot3\cdot2\cdot 1= 120$.)\)

 Apr 6, 2020
 #1
avatar+24983 
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In the prime factorization of \(109!\), what is the exponent of \(3\)?

(Reminder: The number \(n!\) is the product of the integers from \(1\) to \(n\).

For example, \(5!=5\cdot 4\cdot3\cdot2\cdot 1= 120\).)

 

\(\begin{array}{|rcll|} \hline \dfrac{109}{3^1} &=& \lfloor 36.3333333333 \rfloor \\ &=& 36 \\ \hline \dfrac{109}{3^2} &=& \lfloor 12.1111111111 \rfloor \\ &=& 12 \\ \hline \dfrac{109}{3^3} &=& \lfloor 4.03703703704 \rfloor \\ &=& 4 \\ \hline \dfrac{109}{3^4} &=& \lfloor 1.34567901235 \rfloor \\ &=& 1 \\ \hline \dfrac{109}{3^5} &=& \lfloor 0.44855967078 \rfloor \\ &=& 0 \\ \ldots \\ &=& 0 \\ \hline \text{sum} && 36+12+4+1= \mathbf{53} \\ \hline \end{array} \)

 

\(\mathbf{109!} = 2^{104}×\mathbf{3^{\color{red}53}}×5^{25}×7^{17}×11^9×13^8×17^6×19^5×23^4×\cdots\)

 

laugh

 Apr 6, 2020
 #2
avatar+392 
+1

thank u so much

 Apr 6, 2020

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