We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
85
4
avatar

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+5172 
+1

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

 

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

.
 Mar 23, 2019
edited by Rom  Mar 23, 2019
edited by Rom  Mar 23, 2019
 #2
avatar+101391 
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

13 Online Users

avatar
avatar
avatar