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Problem_Set_07_Solutions

# Problem_Set_07_Solutions - From [email protected]

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From [email protected] Thu Jul 12 21:10:08 2001 Date: Fri, 13 Jul 2001 01:01:57 From: Jacob Otto <[email protected]> Subject: discrete math, PS-7 Discrete Math Problem Set 7 1a) 101 99 99 2 101 = 1*99+2 --> 2 = 101 - 1*99 2 1 99 = 49*2+1 ---> 1 = 99 - 49*2 gcd(99,101) = 1 = 99 - 49*2 = 99 - 49 *(101-1*99) = 99 - 49*101 + 49 *99 = 50*99-49*101 which is also equivalent to 50*101-51*99 (x=50, y=49, u=50, v=51) 1b) 35 10 10 5 35 = 3*10+5 --> 5 = 35 - 3*10 5 0 gcd(10,35) = 5 = 1*35-3*10 which is also equivalent to 32*10-9*35 (u=1, w=3, x=32, y=9) 1c) 12 7 7 5 12=1*7+5 --> 5 = 12 - 1*7 5 2 7=1*5+2 --> 2 =7-1*5 2 1 5=2*2+1 --> 1 = 5 -2*2 gcd(7,12) = 1=5 - 2*2 = 5 - 2*(7-1*5) = 3*5 -2*7 = 3*(12-1*7)-2*7=3*12-5*7 which is also equivalent to 7*7-4*12 (u=3,v=5,x=7,y=4) 1d) 42 36 36 6 42=1*36+6 --> 42-1*36 6 0 gcd(36,42) = 6 = 42-1*36 which is also equivalent to 41*36-35*42 (u=1,v=1,x=41,y=35) 2) See printout 3a) (1+x+x^2+x^3+x^4)^3 = (1 + 3x + 6x^2 + 10x^3 + 15x^4 + ...) since the coefficients are those of the 3rd diagonal of Pascal's triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1

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1 5 10 10 5 1 1 6 15 20 15 6 1 x^4 has coefficient 15. 3b) (1+x^2+x^4)^2 * (1+x+x^2)^2 = (1+2x^2+3x^4+2x^6+x^8) * (1 + 2x + 3x^2 + 2x^3 + x^4) Coefficient of x^4: 1+2*3+3 = 10 3c) >From a): (1+x+x^2+x^3+x^4+...) = (1+3x+6x^2+10x^3+15x^4+...)
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