There is 3*5 matrix which can produce signals which in turn help in the illumination of some bulbs. The row of the matrix are denoted by $, & and * from bottom to top and the columns are denoted by the alphabets P, Q, R, S and

T from left to right. $ row contains number which are consecutive multiple of 7, staring from 28 (from left to right). & row contains numbers which are consecutive multiple of 11, starting from 11 (from left to right). * row contains number which are consecutive multiple of 13, starting from 13 (from left to right). The matrix helps in producing signals which can be either a single string of number X- or two-line string X and Y. There are 4 lights A, B, C and D. Based on the outcome of the strings mentioned above one of the light blinks.

For matrix formation:

There is 3 × 5 matrix.

The row of the matrix are denoted by $, & and * from bottom to top and the columns are denoted by the alphabets P, Q, R, S and T from left to right.

$ row contains number which are consecutive multiple of 7, starting from 28 (from left to right).

& row contains number which are consecutive multiple of 11, starting from 11 (from left to right).

* row contains number which are consecutive multiple of 13, starting from 13 (from left to right).

X = *R &S *P *T

X = 39 44 13 65

Condition 2 is applicable here: If an odd number is followed by an even number then the one’s places of all the two-digit numbers are deleted and, tenth place are multiplied to get the outcome

X = 3 × 4 × 1 × 6 = 72

As the outcome is below 85, hence bulb A will blink.

X = $P $R *S &Q

X = 28 42 52 22

Condition 1 is applicable here: If the string has all even numbers, then outcome of the string is obtained by adding all the numbers.

X = 28 + 42 + 52 + 22 = 144

As the outcome of the string is between 111 – 210, hence bulb C will blink.

X = *P &T &P $Q

X = 13 55 11 35

Condition 3 is applicable here: If the string contains 2 prime number, then the tenth’s place is deleted from each of the two-digit number and remaining number are multiplied.

X = 3 × 5 × 1 × 5 = 75

As the outcome is below 85, hence bulb A will blink.

X = $P *Q $S &R

X = 28 26 49 33

Condition 4 is applicable here: If no above logic is followed, then simple outcome is addition of the numbers.

X = 28 + 26 + 49 + 33 = 136

As the outcome of the string is between 111 – 210, hence bulb C will blink.

X = $T *S $S *T

X = 56 52 49 65

Condition 4 is applicable here: If no above logic is followed, then simple outcome is addition of the numbers.

X = 56 + 52 + 49 + 65 = 222

As the outcome of the string is greater than 210, hence bulb D will blink.

Logic:

Number – Subtract the position of 2nd alphabet from right end with 2nd alphabet from left end.

Symbol – Symbol is coded according to number of vowels in the word:

# – 3 @ – 1 * – 4 ! – 5 & – 2

Alphabet – Position of 3rd alphabet from right end. Add same number to it to get the alphabet

For example – Wireless

Number – Position of ‘I’ in English alphabet = 9

Position of ‘S’ in English alphabet = 19

Subtract 19 – 9 = 10

Symbol – Number of vowels in ‘wireless’ = 3

Symbol – #

Alphabet – 3rd alphabet from right end = O

Position of ‘E’ = 5

5 + 5 = 10

Alphabet – J

Hence, ‘wireless’ is coded as ‘J#10’.

Hence, the code for word ‘solution’ is ‘R*0’.

Hence, code for ‘phone call’ is D&6 B@11.

Hence, ‘B#1’ is the code for ‘important’.

Hence, code for ‘delete data’ is ‘J#15 B&19’.

Hence, code for ‘parenting’ is ‘R#13’.

45239 [latex]\Rightarrow[/latex] XFG@# [latex]\Rightarrow[/latex] #F%@#

72163 [latex]\Rightarrow[/latex] !GJB@ [latex]\Rightarrow[/latex] +%+%@

61283 [latex]\Rightarrow[/latex] BJG*@ [latex]\Rightarrow[/latex] @JG*@

85642 [latex]\Rightarrow[/latex] *FBXG

962738 [latex]\Rightarrow[/latex] #BG!@* [latex]\Rightarrow[/latex] *BG!@#

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