If \[17\equiv 3\left( \bmod 7 \right)\] , \[9\equiv 2\left( \bmod 7 \right)\] , then which of the following are true.
 \[17+9=3+2\left( \bmod 7 \right)\]
  \[17+17=9+7\left( \bmod 7 \right)\]
 \[17\times 9=3\times 2\left( \bmod 7 \right)\]
 \[17-9=3-2\left( \bmod 7 \right)\]

(i), (ii) are true
(ii), (iii), (iv) are true
(i), (iii), (iv) are true
All are true

Hint: In the given question, we have given the congruence modulo. In order to solve the question, we need to take one by one and solving each part using the congruence modulo method i.e.
 \[\dfrac{A}{B}=Q\ remainder\ R\] Where A is the dividend, B is the divisor, Q is the quotient and R is the remainder. Therefore, A (mod b) = R.

Complete step-by-step answer:
We have given that,
 \[\Rightarrow 17\equiv 3\left( \bmod 7 \right)\]
This states that 17 is congruent to 3 modulo 7.
As,
17 (mod 7) = 3, so it is in equivalence class for 3.
3 (mod 7) = 3, so it is in equivalence class for 3, as well.
Now,
 \[\Rightarrow 9\equiv 2\left( \bmod 7 \right)\]
This states that 9 is congruent to 2 modulo 7.
As,
9 (mod 7) = 2, so it is in equivalence class for 2.
2 (mod 7) = 2, so it is in equivalence class for 2, as well.
As, we know that,
 \[\dfrac{A}{B}=Q\ remainder\ R\]
Where,
A = dividend
B = divisor
Q = quotient
R = remainder
Thus,
A (mod b) = R
Now we have,
 \[17+9=3+2\left( \bmod 7 \right)\]
17 + 9 = 26
26 = 3 + 23
Here,
We can write 23 as, when we divide 23 by 7 we will get the remainder 2 i.e. 23 (mod 7) = 2
And the value of 2 (mod 7) = 2
Thus,
23 (mod 7) = 2 (mod 7) = 2
So,
26 = 3 + 23 (mod 7) = 3 + 2 (mod 7)
Hence,
 \[17+9=3+2\left( \bmod 7 \right)\]
 \[17+17=9+7\left( \bmod 7 \right)\]
17 + 17 = 34
34 = 9 + 25
Here,
We can write 25 as, when we divide 25 by 7 we will get the remainder 4 i.e. 25 (mod 7) = 4
And the value of 2 (mod 7) = 2
Thus,
25 (mod 7) is not equal to 2 (mod 7)
So, this is not true.
 \[17\times 9=3\times 2\left( \bmod 7 \right)\]
 \[17\times 9=153=3\times 51=3\times 51\left( \bmod 7 \right),\ where\ 51\left( \bmod 7 \right)=2\]
The value of 2 (mod 7) is also = 2
Thus,
51 (mod 7) = 2 (mod 7) = 2
So,
 \[17\times 9=3\times 2\left( \bmod 7 \right)\]
 \[17-9=3-2\left( \bmod 7 \right)\]
 \[17-9=8=3-\left( -5 \right)=3-\left( -5\left( \bmod 7 \right) \right),\ where\ -5\left( \bmod 7 \right)=2\]
The value of 2 (mod 7) = 2
Thus,
-5 (mod 7) is equal to 2 (mod 7)
So,
 \[17-9=3-2\left( \bmod 7 \right)\] .
Therefore,
The option (c ) is the correct answer.
So, the correct answer is “Option C”.

Note: A \[\equiv \] B (mod C) is represented as A is congruent to B modulo C.
The ‘ \[\equiv \] ’ is the symbol for congruence, which represents that the values A and B are in the same equivalence class.
(Mod C) tells us what operation we need to apply to A and B.
When we have both of these that are mentioned in the above two points, we call it as congruence modulo C.