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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 * 4 * 3 * 2* 1 = 120.)

mathbum  Sep 17, 2018
 #1
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+1

Factor 109!:
2^104×3^53×5^25×7^17×11^9×13^8×17^6×19^5×23^4×29^3×31^3×37^2×41^2×43^2×47^2×53^2×59×61×67×71×73×79×83×89×97×101×103×107×109.
These 53 threes come from the following: 
There are 36 multiples of 3s in 109
There are 6 multiples of 3s in multiples of 5
There are 3 multiples of 3s in multiples of 7
There are 3 multiples of 3s in multiples of 11
There are 2 multiples of 3s in multiples of 13
There are 2 multiples of 3s in multiples of 17
There is 1 multiple of 3 in multiples of 19

Guest Sep 17, 2018
 #2
avatar+20591 
+7

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 * 4 * 3 * 2* 1 = 120.)

 

\(\begin{array}{|rcll|} \hline \dfrac{109}{3^1} &=& [36].3333333333 \\\\ \dfrac{109}{3^2} &=& [12].1111111111 \\\\ \dfrac{109}{3^3} &=& [4].03703703704 \\\\ \dfrac{109}{3^4} &=& [1].34567901235 \\\\ \dfrac{109}{3^5} &=& [0].44855967078 \\\\ \ldots &=& [0] \\\\ \hline & \text{sum} =& 36+12+4+1 =\mathbf{ 53} \\ \hline \end{array}\)

 

The exponent of \(\mathbf{3}\) is \(\mathbf{53}\)

 

laugh

heureka  Sep 18, 2018

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