According to the given information,

Number |
0 |
1 |
2 |
3 |
4 |
5 |

Code |
@ |
& |
&@ |
&& |
&@@ |
&@& |

Simplified Code |
0 |
1 |
10 |
11 |
100 |
101 |

So, this entire code language is based on only 2 symbols @ and &, i.e. 0 and 1.

Let’s work out the logic behind the arrangement of these digits.

On observation, we notice that **starting from the rightmost digit**, each digit is multiplied with **corresponding power of 2**, starting with 2^{0}, and then all these terms are added to obtain the number in decimal system.

[The rightmost term is multiplied with 2

^{0}, second term from right is multiplied with 2

^{1}, third term from right is multiplied with 2

^{2} and so on..]

2 → 10 ⇒ (0 × 2^{0}) + (1 × 2^{1}) = (0 × 1) + (1 × 2) = 0 + 2 = 2

3 → 11 ⇒ (1 × 2^{0}) + (1 × 2^{1}) = (1 × 1) + (1 × 2) = 1 + 2 = 3

4 → 100 ⇒ (0 × 2^{0}) + (0 × 2^{1}) + (1 × 2^{2}) = (0 × 1) + (0 × 2) + (1 × 4) = 0 + 0 + 4 = 4

5 → 101 ⇒ (1 × 2^{0}) + (0 × 2^{1}) + (1 × 2^{2}) = (1 × 1) + (0 × 2) + (1 × 4) = 1 + 0 + 4 = 5 and so on.

As per the given question, we need to find the relation between **&@@** and** &@**

&@@ → 100

= (0 × 2^{0}) + (0 × 2^{1}) + (1 × 2^{2})

= 0 + 0 + 4

= 4

&@ → 10

= (0 × 2^{0}) + (1 × 2^{1})

= 0 + 2

= 2

Hence the relation is square root i.e square root (&@@) = &@.

On checking the options one by one:

a) &@&&, &@& → 1011, 101 → 11, 5 → False

b) &&@&, &&& → 1101, 111 → 13, 7 → False

c) &&@&&, &@& → 11011, 101 → 27, 5 → False

d) &&@@&, &@& → 11001, 101 → 25, 5→ True