476**0 represents a six digit number that is divisible by both 3 and 11. If the digits in the hundreds and tens places are non zero and unknown, what are these two digits respectively?

Introduction / Context:
This question tests divisibility rules for 3 and 11 applied to a partially unknown six digit number. The number is written in the pattern 476**0, where the two star positions represent unknown digits. We are told that the digits in the hundreds and tens places are non zero. By translating the divisibility conditions into algebraic relations involving these digits, we can determine their unique values without checking every possibility by trial and error.

Given Data / Assumptions:

- The number has the form 476ab0, where a is the hundreds digit and b is the tens digit. - Digits a and b are non zero and belong to the set {1,2,3,4,5,6,7,8,9}. - The number 476ab0 is divisible by 3 and by 11. - Standard divisibility rules for 3 and 11 apply.

Concept / Approach:
For divisibility by 3, the sum of the digits must be a multiple of 3. For divisibility by 11, the difference between the sum of digits in odd positions and the sum of digits in even positions must be a multiple of 11. We first express these conditions in terms of digits a and b, then solve the resulting modular equations to find a and b. Finally we check which ordered pair matches one of the given options.

Step-by-Step Solution:
Step 1: Write the number as 476ab0 with digits: 4, 7, 6, a, b, 0 from left to right. Step 2: For divisibility by 3, sum of digits is 4 + 7 + 6 + a + b + 0 = 17 + a + b. Step 3: This sum must be a multiple of 3, so 17 + a + b ≡ 0 (mod 3). Since 17 ≡ 2 (mod 3), we get a + b ≡ 1 (mod 3). Step 4: For divisibility by 11, label positions from left: A = 4, B = 7, C = 6, D = a, E = b, F = 0. Step 5: Sum of digits in odd positions is A + C + E = 4 + 6 + b = 10 + b. Step 6: Sum of digits in even positions is B + D + F = 7 + a + 0 = 7 + a. Step 7: Condition for divisibility by 11 is (10 + b) − (7 + a) ≡ 0 (mod 11), which simplifies to 3 + b − a ≡ 0 (mod 11), so b − a ≡ 8 (mod 11). Step 8: Now a and b are digits from 1 to 9. Solving b − a ≡ 8 (mod 11) within this range gives possible pairs (1, 9), (5, 2) and (8, 5). Step 9: Check the condition a + b ≡ 1 (mod 3) for each pair. Only (8, 5) gives a + b = 13, and 13 ≡ 1 (mod 3). Step 10: Therefore, the hundreds digit is 8 and the tens digit is 5.

Verification / Alternative check:
Form the number 476850 and verify quickly. Sum of digits is 4 + 7 + 6 + 8 + 5 + 0 = 30, which is divisible by 3. For divisibility by 11, sum of digits at odd positions is 4 + 6 + 5 = 15, and sum at even positions is 7 + 8 + 0 = 15. Their difference is 15 − 15 = 0, which is a multiple of 11. Hence 476850 is divisible by both 3 and 11, confirming the correctness of the chosen digits.

Why Other Options Are Wrong:
For (6 and 2), (8 and 2) and (6 and 5), either the sum of digits is not a multiple of 3 or the alternating sum for 11 is not a multiple of 11. Detailed checking shows that none of those combinations satisfy both divisibility conditions simultaneously, so they must be discarded.

Common Pitfalls:
Some learners may apply divisibility rules only for 3 or only for 11, choosing pairs that satisfy one condition but not the other. Another common mistake is misplacing the digits when computing alternating sums for 11 or forgetting to include the zero in the units place. Working systematically with modular arithmetic and carefully listing odd and even positions avoids these errors.

Final Answer:
The non zero digits in the hundreds and tens places are 8 and 5 respectively, which matches option D.