What is the largest integer n such that 3^n is a factor of 1*3*5* ...*97*99*101*... *197*199?
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
+23252 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.
+118687 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 :)
