Problem 7. The symbol n! is read "n factorial" and means the product of the first n positive integers: 1 · 2 · . . . · (n - 2) · (n - 1) · n. Let p be the positive integer which is the largest power of 3 which divides evenly into 100! (100 factorial). That is, divides into 100! leaving no remainder, but has a remainder when it is divided into 100! Find p. [Problem submitted by Don Hentschel, LACC Assistant Professor of Mathematics.] Solution: a) 100! = 1 · 2 · 3 · . . . · 98 · 99 · 100 b) 3 divides evenly into every third factor: 3, 6, 9, ..., 99. There are 33 of these. c) 3 divides evenly again into every ninth factor: 9, 18, 27,..., 99. There are 11 of these. d) 3 divides evenly again into every twenty-seventh factor: 27, 54, 81. There are 3 of these. e) 3 divides evenly again into every eighty-first factor: 81. There is one of these. f) So, 3 divides evenly into 100! 33 + 11 + 3 + 1 = 48 times. Therefore, p = 48.

| Problem 1 | Problem 2 | Problem 3 | Problem 4 | Problem 5 |
| Problem 6 | Problem 7 | Problem 8 | Problem 9 | Problem 10 |

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- Last Updated: 8/29/16