Question 4:
4 | x = 3a + 1 = 11b + 5
mod 11: 3a  4  15  a  5 (mod 11) (since hcf(3, 11) = 1)
(a = 11c + 5)
x = 33c + 16
2x = 66c + 32 = 17d + 5 mod 17: 15c  –27  5c  –9  8  25
 c  5 (mod 17) (since hcf(5, 17) = 1)
Then c = 17e + 5  x = 561e + 181 or x  181 (mod 561)
Stating at least one of either hcf(3,11) = 1 or hcf(5,17)=1 | M1
M1
A1
M1
A1
B1 | 1.1
2.1
3.1a
2.1
2.2a
2.4 | Equating first two equations
Working mod 11
or via b = 3c + 1 (working mod 3)
Substituting into 3rd equation and working mod 17
Alternative Method 1
2x  5  22  x  11 (mod 17)  –6 (mod 17) | M1
Then x  –6 (mod 11) | M1
A1
x  –6 (mod 187)  181
since hcf(11, 17) = 1 | B1
Note that 181 is already  1 (mod 3) | M1 | OR continue as above: x = 3a + 1 = 187b + 181
Since there is a unique solution (mod 3  11  17 = 561) | (as 3, 11, 17 are pairwise co-prime)
the solution is x  181 (mod 561) | A1
Alternative Method2 The Chinese Remainder Theorem (CRT)
There is a unique solution mod M = 3  11  17 = 561 | B1
2x  5  22  x  11 (mod 17) | M1
M
a = 1, M = = 1 8 7 , R is the reciprocal of M mod 3 = 1
1 1 1 1
3 | M1 | Since 187  1 (mod 3) anyway
For any one set of terms correct
M
a = 5, M = = 5 1 , R is the reciprocal of M mod 11 = 8
2 2 2 2
1 1 | Since 51  7 (mod 11) and 7  8 = 56  1 (mod 11)
M
a = 11, M = =33, R is the reciprocal of M mod 17 = 16
3 3 3 3
17 | A1 | All three sets of terms correct. Since 33  –1  16
(mod 17)
Then x  a M R + a M R + a M R (mod 561)
1 1 1 2 2 2 3 3 3 | M1 | CRT employed correctly (in principle)
= 11871 + 5518 + 113316 = 8035  181 (mod 561) | A1 | BC
[6]
M1
A1