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avatar+188 

How many even natural-number factors does n = (2^2)(3^1)(7^2) have?

 

ok so i think that all the odd factors are 3, 21, 150, 49, but i'm not sure about the rest :(

 Jan 9, 2020
 #1
avatar+19815 
+3

4 x 3 x 49 = 588

Factors (from a website):

1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 49, 84, 98, 147, 196, 294, 588,
-1, -2, -3, -4, -6, -7, -12, -14, -21, -28, -42, -49, -84, -98, -147, -196, -294, -588     Pick the ones you like and count them up !

 Jan 9, 2020
 #2
avatar+188 
+1

oh thats smart.. thank you!!

atlas9  Jan 9, 2020
 #3
avatar+106519 
+2

Here's  a way to figure this

 

2^1 = 2

2^2 = 4

2 * 3  =  6

2^2 * 3  = 12

2* 7   =  14

2^2 * 7 = 28

2 * 3 * 7  = 42

2^2 *3 * 7  = 84

2*7^2 =  98

2^2 * 7^2  = 196

2 *3 * 7^2  =  294

2^2 * 3 * 7^2 =  588

 

[ Note....negative factors  are not "natural"  numbers ]

 

 

 cool cool cool

 Jan 9, 2020
 #4
avatar+188 
+1

WOAH thats smart too thanks CPhill!!

atlas9  Jan 9, 2020
 #5
avatar+23812 
+4

How many even natural-number factors does \(n = (2^2)(3^1)(7^2)\) have?

 

\(\begin{array}{|llcll|} \hline 2^{\color{red}2} * & 3^{\color{blue}1} * 7^{\color{green}2} \\\\ ({\color{red}2}+1 ) *& ({\color{blue}1}+1) * ({\color{green}2}+1) = 3*2*3 &=& 18~ \text{divider} \\ & ({\color{blue}1}+1) * ({\color{green}2}+1) = 2*3 &=& 6~ \text{odd divider} \qquad (\text{odd*odd=odd})\\ \hline &&& 18-6 = \mathbf{12}~ \text{even divider} \\ \hline \end{array} \)

 

laugh

 Jan 9, 2020
 #6
avatar+188 
+1

Thank you all so much for your answers, they helped me a lot!! :))

 Jan 10, 2020

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