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When all of the factors of 90 are multiplied together, the product can be expressed as 90^k. What is the value of k? 

 Mar 23, 2019
 #1
avatar+6192 
+1

\(\text{listing the divisors of 90}\\ 1,2,3,5,6,9,10,15,18,30,45,90\\ \text{We can collect pairs of divisors that multiply to 90}\\ (1,90),(2,45),(3,30),(5,18),(6,15),(9,10)\\ \text{We see then that the product is }90^6\)

 

\(\text{In general the product of the factors of }n \text{ will be }n^{\frac 1 2\text{# of factors of n}}\\ \text{You can determine the # of factors of }n \text{ by taking the product of each of the}\\ \text{exponents of the prime factors plus 1. For example }\\ 90 = 2^1\cdot 3^2 \cdot 5^1\\ \text{It has }(1+1)(2+1)(1+1) = 12 \text{ factors which multiply to }90^{12/2} = 90^6\)

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 Mar 23, 2019
edited by Rom  Mar 23, 2019
edited by Rom  Mar 23, 2019
 #2
avatar+111392 
0

Factors of 90 =

 

 1 2 3 5  6 9 10 15 18 30  45  90

 

We  can write

 

(1 * 90) * (2* 45 )  *  (3 * 30)  * (5 * 18)   * (6 * 15 )  * ( 9 *10)  =

 

   90    *     90        *      90      *     90      *     90        *   90     =

 

90^6

 

k  =  6

 

cool cool cool

 Mar 23, 2019
 #4
avatar+533 
0

We know that if \(x\) is a factor of \(n\), then \(\frac{n}{x}\) is also a factor of \(n\). Because \(90\) is not a perfect square, we can pair up the \(x\)'s and the \(\frac{n}{x}\)'s and multiply them to get \(n\). Because this is true for every number, we can count the number of factors of \(90 = 2 \cdot 3^2 \cdot 5\). We can choose 2 ways for the exponent of 2, 3 for the exponent of 3, and 2 for the exponent of 5. So, there are \(2 \cdot 3 \cdot 2 = 12\) factors. This means there are \(\frac{12}{2}=6\) paris that multiply to \(90\), and their products together is \(90^6\), so the answer is \(\boxed{k=6}\).

Hoping this helped, 

asdf334

 Mar 23, 2019

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