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Solutions_MT1sum11

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

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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 ϕ (9) = 3 2 - 3 1 = 6 . By Euler’s 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

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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 . Answer: p + 1. Solution: We have to count all multiples of p that appear in the defining product for ( p 2 )! = 1 · 2 · . . . · p 2 , that is, all multiples of p between 1 and p 2 . These are p, 2 p, . . . , ( p - 1) p and p 2 , the last one contributes to the second power of
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Solutions_MT1sum11 - MAT 312/AMS 351 Summer 2011 Solutions...

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