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Solutions_MT1s11

# Solutions_MT1s11 - MAT 312/AMS 351 Spring 2011 Solutions to...

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MAT 312/AMS 351, Spring 2011 Solutions to the Midterm I Maximal grade: 100 points 1. (15 points) Find the prime decomposition of the number 10! = 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 . Answer: 10! = 2 8 · 3 4 · 5 2 · 7 . Solutions: We can consider the prime decomposition of every integer between 1 and 10 and gather the factors together: 10! = 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 · 10 = 1 · 2 · 3 · 2 2 · 5 · (2 · 3) · 7 · 2 3 · 3 2 · (2 · 5) = 2 8 · 3 4 · 5 2 · 7 . Grading: -5 points for the incorrect or missing simplification, -1 point if 1 is included in the prime decomposition 2. (20 points) Find all n such that 2 n + 3 n is divisible by 5. Answer: n is odd. Solutions: a) Let us describe powers of 2 and 3 modulo 5. We have 2 2 = 4 , 2 3 = 8 = 3 , 2 4 = 2 · 3 = 1 mod 5 , 3 2 = 9 = 4 , , 3 3 = 3 · 4 = 12 = 2 , 3 4 = 2 · 3 = 1 mod 5 . Therefore the pattern of 2 n mod 5 is 2 , 4 , 3 , 1 , 2 , 4 , 3 , 1 . . . 1

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and the pattern of 3 n mod 5 is 3 , 4 , 2 , 1 , 3 , 4 , 2 , 1 . . . , so the pattern of 2 n + 3 n mod 5 is 0 , 3 , 0 , 2 , 0 , 3 , 0 , 2 . . . If 2 n + 3 n is divisible by 5, then 2 n + 3 n = 0 mod 5, and this happens for odd values of n . b) Remark that 3 = - 2 mod 5, so 3 n = ( - 1) n 2 n mod 5. Therefore for odd n we have 3 n = - 2 n mod 5, so 2 n + 3 n = 0 mod 5, and for even n we have 3 n = 2 n mod 5, so 2 n + 3 n = 2 · 2 n 6 = 0 mod 5.
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