Solutions_MT1sum11

Solutions_MT1sum11 - MAT 312/AMS 351, Summer 2011 Solutions...

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Unformatted text preview: MAT 312/AMS 351, Summer 2011 Solutions to the Midterm I Maximal grade: 100 points 1. (15 points) Find the last digit of the number 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! + 10! Answer: 3 . Solution: We have 1! = 1 , 2! = 2 , 3! = 6 , 4! = 24 . For n 5 the number n ! is divisible by 2 and by 5, so it is divisible by 10, and its last digit is equal to 0. Therefore 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! + 10! = 1 + 2 + 6 + 4 = 3 mod 10 . 2. (20 points) Compute 17 25 mod 9. Answer: 17 25 = 8 mod 9. Solutions: a) Since 9 = 3 2 , we have (9) = 3 2- 3 1 = 6 . By Eulers theorem (we can use it since g.c.d (17 , 9) = 1) we have 17 6 = 1 mod 9 . Therefore 17 25 = (17 6 ) 4 17 = 1 4 8 = 8 mod 9 . b) Remark that 17 = 8 mod 9, so 17 2 = 8 2 = 64 = 1 mod 9 . 1 Therefore 17 25 = (17 2 ) 12 17 = 1 12 8 = 8 mod 9 . 3. (10 points) Let p be a prime number. Find the maximal power of p that divides ( p 2 )! Prove that your answer is true for all prime p ....
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This note was uploaded on 01/25/2012 for the course MAT 312 taught by Professor Bescher during the Summer '08 term at SUNY Stony Brook.

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Solutions_MT1sum11 - MAT 312/AMS 351, Summer 2011 Solutions...

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