QUESTION

CSAT

Hard

Reasoning

Prelims 2022

An Identity Card has the number ABCDEFG, not necessarily in that order, where each letter represents a distinct digit $(1, 2, 4, 5, 7, 8, 9 \text{ only})$ . 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?

Select an option to attempt

Explanation

To solve the problem involving the 7-digit number ABCDEFG and find the value of $C + D + E$ (the sum of the middle three digits), we can apply divisibility rules effectively without needing to solve for all variables.

Given Conditions and Steps:

Divisibility by 6: The number ABCDEF (after deleting G) is divisible by 6, meaning it must be divisible by both 2 and 3. Since it must be divisible by 2, F must be an even digit (2, 4, or 8). For divisibility by 3, the sum of the digits $A + B + C + D + E + F$ must be divisible by 3.
Last Digit Condition: G must be 9 to ensure ABCDEF is divisible by 3.
Divisibility by 5: After removing F and G (giving ABCDE), ABCDE must be divisible by 5, which implies $E = 5$ .
Divisibility by 4: Removing 3 digits from the right gives ABCD, which must be divisible by 4. Therefore, D can only be 2 or 4 (the last two digits of ABCD need to form a number divisible by 4).
Divisibility by 2: When AB is examined (after removing 5 digits), it must also be divisible by 2, which means B must be even (2, 4, or 8).

Key Inferences: Since F, D, and B are even digits, we have three potential values for each. A and C can only be 1 or 7 (the remaining odd digits), based on the constraints.

Divisibility by 3: For ABC to be divisible by 3, the sum $A + B + C$ must be divisible by 3. With B set to 4 (the only even digit that allows $A + B + C$ to remain under 12 while meeting other conditions), we can conclude $A + 4 + C \equiv 0 \mod 3$ .

Conclusion: Since D can only be 2 (given the analysis above) and with $E = 5$ , we can conclude: Let $C = 1$ and $D = 2$ . Thus, $C + D + E = 1 + 2 + 5 = 8$ .