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NUMBER SENSE (PROBLEM SOLVING)

Discover the fascinating world of number theory by exploring how many consecutive zeros are at the end of 100!. This problem-solving challenge analyzes the prime factorization of 100! to determine the number of trailing zeros, which is equal to the number of 5s in its factors. We also delve into evaluating remainders, counting positive integers less than 70 that are relatively prime to 70, and discuss activities from the Alabama Statewide Mathematics Contest. Enhance your understanding and skills in mathematics!

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NUMBER SENSE (PROBLEM SOLVING)

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  1. NUMBER SENSE(PROBLEM SOLVING) TIMES3_ Number theory Day5

  2. Factoring Formulas

  3. Inspired by…

  4. How many consecutive zeros does 100! have at the end? Assume that 100! has m 2s and n 5s in its prime factorization. Apparently, there are more 2s than 5s in 100!. Thus, the number of zeros at the end of 100! is same as the number of 5s in 100! Let’s count: 100!= # of multiples of 5 less than equal to 100 = # of multiples of 25 less than equal to 100 = Answer:

  5. What is the remainder if is divided by ?

  6. Find the remainder when the sum is divided by .

  7. Let . Evaluate .

  8. How many positive integers less than 70 are relatively prime to 70?

  9. Alabama Statewide Mathematics Contest by ACTM • You can go to http://mcis.jsu.edu/mathcontest/and check the contest problems with answers for last 12 years.

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