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If 2 is added to each even digit and 3 is subtracted from each odd digit in the number 9472561, what is the sum of the digits which are third from the left and third from the right?
7
9
8
6
7
Identify only the 3rd digit from the left and 3rd digit from the right; do not waste time transforming the entire number.
Identify only the 3rd digit from the left and 3rd digit from the right; do not waste time transforming the entire number.
Identify Target Digits
Original number is 9472561. The 3rd digit from the left is 7. The 3rd digit from the right is 5.
Apply Transformation Rules
Rule: If odd, subtract 3; if even, add 2. For the 3rd digit from left (7): 7 is odd, so 7 - 3 = 4. For the 3rd digit from right (5): 5 is odd, so 5 - 3 = 2.
Calculate Final Sum
The transformed digits are 4 and 2. Sum = 4 + 2 = 6. Wait, re-evaluating indices: Left (1:9, 2:4, 3:7), Right (1:1, 2:6, 3:5). 3rd from left is 7 (odd тЖТ 7-3=4). 3rd from right is 5 (odd тЖТ 5-3=2). Sum is 6. Re-checking original question logic: 9472561. Left 3rd is 7, Right 3rd is 5. 7 is odd (7-3=4), 5 is odd (5-3=2). Sum = 6. Let me re-read: 3rd from left is 7. 3rd from right is 5. Sum is 6. If the question implies Option A is 7, perhaps the 3rd digit from the right is 6? No, 6 is 2nd from right. Perhaps 2 is 3rd? No. Let's re-calculate: 9(1), 4(2), 7(3), 2(4), 5(5), 6(6), 1(7). 3rd from left is 7. 3rd from right (1, 6, 5) is 5. 4 + 2 = 6.
B is wrong because it results from incorrect digit selection, C is wrong because it represents a sum miscalculation, D is correct based on the logic but the provided key indicates 7, likely implying a different digit mapping or counting index interpretation. Assuming standard positional counting, 6 is the mathematically sound result.
D is correct because the 3rd digit from the left is 7 (odd, so 7-3=4) and the 3rd digit from the right is 5 (odd, so 5-3=2), giving a sum of 6.
In sequence problems, always define the direction (left/right) first to avoid off-by-one errors.