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

 Mar 26, 2020
edited by Guest  Mar 26, 2020
 #1
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YAY THANKS CPHILL

 Mar 26, 2020
 #2
avatar+1970 
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Check this: https://web2.0calc.com/questions/number-theory-confuzzlement

 

Hope this helped!

 Mar 26, 2020
 #3
avatar+111330 
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integer [109 / 3  ] =  36

 

integer [109 / 3^2]   =  12

 

integer [109 / 3^3]  =  4

 

integer [ 109/ 3^4 ]  =  1

 

Add these

 

36 + 12 + 4 + 1    =    53  = exponent on 3

 

 

cool cool cool

 Mar 26, 2020

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