hw03sol - Math 55: Discrete Mathematics, Fall 2008 Homework...

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Unformatted text preview: Math 55: Discrete Mathematics, Fall 2008 Homework 3 Solutions 3.5: 6. There are 24 zeroes at the end of 100!. To see this, let e be the exponent of 2 and f exponent of 5 in the prime factorization of 100!. Then the largest power of 10 that divides 100! is 10 min( e,f ) . To calculate f , observe that 20 of the factors in 100! = 1 2 100 are multiples of 5, and of those, four (25, 50, 75, and 100) contain an extra factor of 5. Since none of them contains a factor 5 3 , we get f = 24. Since there are 50 even factors, we see that e 50, and therefore the minimum of e and f is 24, without further calculating e . *18. We can assume n > 1, so n is not relatively prime to itself. Therefore ( n ) is equal to the number of members of the set { 1 , 2 ,...,n- 1 } that are relatively prime to n , and thus ( n ) = n- 1 if and only if every positive integer less than n is relatively prime to n . If n is prime, then clearly this condition holds. Conversely, if the condition holds, then n has no divisor 1 < d < n , since then...
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This note was uploaded on 08/16/2010 for the course MATH 55 taught by Professor Strain during the Fall '08 term at University of California, Berkeley.

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hw03sol - Math 55: Discrete Mathematics, Fall 2008 Homework...

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