Maths Olympiad Prep

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Algebra Difficulty 5.6 AIME, harder Find the answer

9. Use the results of problem 8 to find all solutions of each of the following systems
a)
x+y≡1( mod 7)x+z≡2( mod 7)y+z≡3( mod 7)\begin{array}{l} x+y \equiv 1(\bmod 7) \\ x+z \equiv 2(\bmod 7) \\ y+z \equiv 3(\bmod 7) \end{array}
b)
x+2y+3z≡1( mod 7)x+3y+5z≡1( mod 7)x+4y+6z≡1( mod 7)\begin{array}{l} x+2 y+3 z \equiv 1(\bmod 7) \\ x+3 y+5 z \equiv 1(\bmod 7) \\ x+4 y+6 z \equiv 1(\bmod 7) \end{array}
c)
x+y+z≡1( mod 7)x+y+w≡1( mod 7)x+z+w≡1( mod 7)y+z+w≡1( mod 7)\begin{array}{l} x+y+z \equiv 1(\bmod 7) \\ x+y+w \equiv 1(\bmod 7) \\ x+z+w \equiv 1(\bmod 7) \\ y+z+w \equiv 1(\bmod 7) \end{array}

A number or a short expression. Spacing and $ signs are ignored.

Solution

9. a) x≡0,y≡1,z≡2( mod 7)x \equiv 0, y \equiv 1, z \equiv 2(\bmod 7)
c) x≡5,y≡5,z≡5,w≡5( mod 7)x \equiv 5, y \equiv 5, z \equiv 5, w \equiv 5(\bmod 7)

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.