Q. 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.

Explanation : 

An Identity Card has the number ABCDEFG

The number ABCDEFG is divisible by 9. Since the sum of the digits of a number divisible by 9 is also divisible by 9, we know that A+B+C+D+E+F+G = 1+2+4+5+7+8+9 = 36.

After deleting the first digit from the right (G), the resulting number ABCDEF is divisible by 6. This means that F must be even (2 or 8) and the sum of the digits (A+B+C+D+E+F) must be divisible by 3.

After deleting two digits from the right (FG), the resulting number ABCDE is divisible by 5. This means that E must be 5.

After deleting three digits from the right (EFG), the resulting number ABCD is divisible by 4. For a number to be divisible by 4, the last two digits must form a number divisible by 4. So, CD must be one of the following pairs: 12, 24, 28, 52, 72, 92.

After deleting four digits from the right (EFGD), the resulting number ABC is divisible by 3. This means that the sum of A, B, and C must be divisible by 3.

After deleting five digits from the right (EFGDC), the resulting number AB is divisible by 2. This means that B must be even (2 or 8).

Now let’s use this information to find the possible value for the sum of the middle three digits (C, D, and E).

Since E is 5, and B is even (2 or 8), we can eliminate the pairs 52 and 92 for CD, as they would require B to be odd. We are left with these possibilities for CD: 12, 24, 28, 72.

Consider CD = 12. Then, A = 7, B = 8, C = 1, D = 2, E = 5, F = 4, and G = 9. The conditions for divisibility by 9, 6, 5, 4, 3, and 2 are satisfied with 7815249. The sum of the middle three digits (C, D, and E) is 1+2+5 = 8.

In this case, the correct answer is (a) 8.