12/13/2017
OCR Further Maths Additional Pure Book
In the above example we could have started with x ≡ 4(mod 7) ⇒ x = 4 + 7k and then written this in modulo 5 and proceeded in the same way.
Tip
WORKED EXAMPLE 2.14
PL
Solve the simultaneous linear congruences:
E
Most errors occur when solving the linear congruences, e.g. 5k ≡ 1(mod 7) should give k ≡ 3(mod 7). Always check this answer before completing the problem.
x ≡ 4(mod 6) …(i)
x ≡ 2(mod 5) …(ii)
x ≡ 2(mod 4) …(iii)
Where k is an integer.
Write x ≡ 4 + 6k ≡ 2(mod 5) ∴ 6k ≡ − 2 ≡ 3(mod 5) ∴ k ≡ 3(mod 5)
Equating (i) and (ii) and using the modulo of (ii). Simplify and solve.
SA M
(i) is x ≡ 4(mod 6) ⇒ x = 4 + 6k…( * )
Hence k = 3 + 5m (m is an integer) and substituting in (*) x = 4 + 6(3 + 5m) = 22 + 30m Hence the solution to (i) and (ii) is: x ≡ 22(mod 30). …(iv)
Now solve the simultaneous equations (iii) and (iv): (iii) ⇒ x ≡ 2(mod 4) ∴ x = 2 + 4k and x ≡ 2 + 4k ≡ 22(mod 30) 4k ≡ 20(mod 30) ⇒ 2k ≡ 10(mod 15) ⇒ k ≡ 5(mod 15)
Note – change of modulo since 4k ≡ 20(mod 30) ⇒ 4k = 20 + 30t ÷ 2 ⇒ 2k = 10 + 15t ∴ 2k ≡ 10(mod 15)
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