Direction: In a certain number system there are only two notations to...
According to the given information,
So, this entire code language is based on only 2 symbols % and #, i.e. 0 and 1.
Logic:
On observation, we notice that starting from the rightmost digit, each digit is multiplied with corresponding power of 2, starting with 20, and then all these terms are added to obtain the number in the decimal system.
[The rightmost term is multiplied with 20, the second term from right is multiplied with 21, the third term from right is multiplied with 22 and so on..]
2 → 10 ⇒ (0 × 20) + (1 × 21) = (0 × 1) + (1 × 2) = 0 + 2 = 2
3 → 11 ⇒ (1 × 20) + (1 × 21) = (1 × 1) + (1 × 2) = 1 + 2 = 3
4 → 100 ⇒ (0 × 20) + (0 × 21) + (1 × 22) = (0 × 1) + (0 × 2) + (1 × 4) = 0 + 0 + 4 = 4
5 → 101 ⇒ (1 × 20) + (0 × 21) + (1 × 22) = (1 × 1) + (0 × 2) + (1 × 4) = 1 + 0 + 4 = 5 and so on.
Based on this logic,
‘###%#’ ↔ 11101
11101 ⇒ (1 × 20) + (0 × 21) + (1 × 22) + (1 × 23) + (1 × 24)
= (1 × 1) + (0 × 2) + (1 × 4) + (1 × 8) + (1 + 16)
= 1 + 0 + 4 + 8 + 16 = 29
‘#%’ ↔ 10
10⇒ (0 × 20) + (1 × 21)
= (0 × 1) + (1 × 2)
= 0 + 2 = 2
'##%%' ↔ 1100
1100 ⇒ (0 × 20) + (0 × 21) + (1 × 22) + (1 × 23)
= 0 + 0 + 4 + 8 = 12
'##%' ↔ 110
110 ⇒ (0 × 20) + (1 × 21) + (1 × 22)
= 0 + 2 + 4 = 6
'#%#' ↔ 101
101 ⇒ (1 × 20) + (0 × 21) + (1 × 22)
= 1 + 0 + 2 = 5
The resultant of the expression is: 29 ÷ 2 × 12 + 6 × 5 = 204
Now, we need to convert 204 into the given code language.
204 = 128 + 64 + 8 + 4
∴ 204 = (0 × 20) + (0 × 21) + (1 × 22) + (1 × 23) + (0 × 24) + (0 × 25) + (1 × 26) + (1 × 27)
∴ 204 ⇒ 11001100
Converting 0s and 1s to %s and #s respectively, we get
729 = #%##%##%%#
So, the square of "##%##" is "#%##%##%%#".
Hence, the correct option is (B).