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Let $k$ be the largest integer such that $\dfrac{343!}{7^k}$ is an integer. What is the remainder when $\dfrac{343!}{7^k}$ is divided by $7$?

 Sep 23, 2020
 #1
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343!= 2^337 * 3^169 * 5^83 * 7^57 * 11^33 * 13^28 * 17^21 * 19^18 * 23^14 * 29^11 * 31^11 * 37^9 * 41^8 * 43^7 * 47^7 * 53^6 * 59^5 * 61^5 * 67^5 * 71^4 * 73^4 * 79^4 * 83^4 * 89^3 * 97^3 * 101^3 * 103^3 * 107^3 * 109^3 * 113^3 * 127^2 * 131^2 * 137^2 * 139^2 * 149^2 * 151^2 * 157^2 * 163^2 * 167^2 * 173 * 179 * 181 * 191 * 193 * 197 * 199 * 211 * 223 * 227 * 229 * 233 * 239 * 241 * 251 * 257 * 263 * 269 * 271 * 277 * 281 * 283 * 293 * 307 * 311 * 313 * 317 * 331 * 337.

 

The largest k = 57 and [343! / 7^57] mod 7 = 6

 Sep 23, 2020
 #2
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How did you get [343! / 7^57] mod 7 = 6? Why does it equal 6?

Guest Sep 23, 2020

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