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What is the largest integer n such that 3^n is a factor of 1*3*5* ...*97*99*101*... *197*199?

 Mar 2, 2022
 #1
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+1

199!! [double factorial] == 3^49 × 5^25 × 7^16 × 11^10 × 13^9 × 17^6 × 19^5 × 23^4 × 29^3 × 31^3 × 37^3 × 41^2 × 43^2 × 47^2 × 53^2 × 59^2 × 61^2 × 67 × 71 × 73 × 79 × 83 × 89 × 97 × 101 × 103 × 107 × 109 × 113 × 127 × 131 × 137 × 139 × 149 × 151 × 157 × 163 × 167 × 173 × 179 × 181 × 191 × 193 × 197 × 199.

 

As you can see, the largest n==49

 Mar 2, 2022
 #2
avatar+23198 
+1

I have a different answer:

 

199 / 3  =  66.3333... which to me indicates that there are 66 numbers

between 1 and 99 that each has a factor of 3     --->     66

 

199 / 9  =  22.1111... which to me indicaets that there are 22 numbers

between 1 and 99 that each has a second factor of 3     --->     22

 

199 / 27  =  7.370... which to me indicates that there are 7 numbers

between 1 and 99 that each has a third factor of 3     --->     7

 

199 / 81  =  2.456... which to me indicates that there are 2 numbers

between 1 and 99 that each has a fourth factor of 3     --->     2

 

Therefore, i arrive at 66 + 22 + 7 + 2  =  97 factors of 3.

 Mar 2, 2022
 #3
avatar+117834 
+1

Geno, your answer would be great if it was 199! but it isn't.  All the even numbers are missing.

 

odd multiples of 3,    6n-3  from n=1 to   n=round down[(199+3)/6] = 33          33 of them

odd multiples of 9      9,27,.....189                 9*1 to 9*21                                  11  of them

odd multiples of 27     27  ,...  179                 27*1 to 27*7                                        4 of them

odd multiples of 81     81  that is the only one                                                          1 of them

 

33+11+4+1 = 49

 

So guest and I agree.  But i used Genos method  :)

Melody  Mar 2, 2022
edited by Melody  Mar 2, 2022
edited by Melody  Mar 2, 2022
 #4
avatar+23198 
+1

Thanks, Melody ... I misread it ....

 Mar 2, 2022
 #5
avatar+117834 
0

Yes, not big deal.

I liked your approach  :)

Melody  Mar 3, 2022

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