An Identity Card has the number ABCDEFG, not necessarily in that order, where each letter represents a distinct digit (1, 2, 4, 5, 7, 8, 9 only). The number is divisible by 9. After deleting the first digit from the right, the resulting number is divisible by 6. After deleting two digits from the right of original number, the resulting number is divisible by 5. After deleting three digits from the right of original number, the resulting number is divisible by 4. After deleting four digits from the right of original number, the resulting number is divisible by 3. After deleting five digits from the right of original number, the resulting number is divisible by 2.

Which of the following is a possible value for the sum of the middle three digits of the number?

Divisibility Rule of 2 – If a number is even or a number whose last digit is an even number i.e. 2,4,6,8 including 0, it is always completely divisible by 2.

Divisibility Rule of 3 – A number is divisible by 3 if the sum of its digits is divisible by 3

Divisibility Rule of 5 – Numbers, which last with digits, 0 or 5 are always divisible by 5.

The 7 digit number is ABCDEFG and we have to find the value of C + D + E (middle 3 digits)

After deleting 1 digit from the right, the resulting number (ABCDEF) is divisible by 6. Hence by divisibility rule of 2, F = 2, 4 or 8 (i.e. an even number).

The remaining number (ABCDEF) is divisible by 3, then G must be 9 to satisfy the divisibility test of 3.

After deleting 2 digits from the right, the resulting number (ABCDE) is divisible by 5. It means that, E = 5. Now, C+D can be between 3 to 7 (as per 4 choices given)

After deleting 3 digits from the right, the resulting number (ABCD) is divisible by 4. It means that, D = 2 or 4 (as per divisibility test and 8 can be dropped due to above condition)

After deleting 5 digits from the right, the resulting number (AB) is divisible by 2. It means that, B = 2, 4 or 8 (i.e. an even number).

So, F, D and B are even numbers (2, 4 or 8). Therefore, A & C are 1 or 7 (only two remaining choices.

After deleting 4 digits from the right, the resulting number (ABC) is divisible by 3, so now as per the divisibility rule, B must be 4 (A & C are already known as 1 or 7, A+B+C must be divisible by 3)

Now, C & D cannot be 7 and 8 as then C+E or D+E will exceed highest option 12, so won’t fit any of the options. So, C = 1 and D = 2

So, C+D+E = 1+2+5 = 8