(a)     Find the general solution for the following system of congruences.

     $N \equiv 3(\bmod 11)$

     $N \equiv 4(\bmod 9)$

     $N \equiv 0(\bmod 7)$

(b)     Find all values of N such that $2000 \leqslant N \leqslant 4000$.

(a)     $N = 3 + 11t$     M1

$3 + 11t \equiv 4(\bmod 9)$

$2t \equiv 1(\bmod 9)$     (A1)

multiplying by 5, $10t \equiv 5(\bmod 9)$     (M1)

$t \equiv 5(\bmod 9)$     A1

$t = 5 + 9s$     M1

$N = 3 + 11(5 + 9s)$

$N = 58 + 99s$     A1

$58 + 99s \equiv 0(\bmod 7)$

$2 + s \equiv 0(\bmod 7)$

$s \equiv 5(\bmod 7)$     A1

$s = 5 + 7u$     M1

$N = 58 + 99(5 + 7u)$

$N = 553 + 693u$     A1

Note: Allow solutions that are done by formula or an exhaustive, systematic listing of possibilities.

[9 marks]

(b)     u = 3 or 4

hence N = 553 + 2079 = 2632 or N = 553 + 2772 = 3325     A1A1

[2 marks]

Total [11 marks]

This was a standard Chinese remainder theorem problem that many candidates gained good marks on. Some candidates employed a formula, which was fine if they remembered it correctly (but not all did), although it did not always show good understanding of the problem.