In the six-digit number 4760, the two asterisks represent unknown non-zero digits in the hundreds and tens places respectively. If the complete number 4760 is divisible by both 3 and 11, what are these two digits (hundreds digit and tens digit) respectively?

Introduction:
This divisibility question combines two classic rules: divisibility by 3 and divisibility by 11. You are given a six-digit number with two missing digits and asked to determine those digits so that the resulting number is divisible by both 3 and 11, with the additional condition that the missing digits are non-zero.

Given Data / Assumptions:

• The number has the form 4 7 6 a b 0, where a is the hundreds digit and b is the tens digit.
• Digits a and b are non-zero digits from 1 to 9.
• The complete number 476ab0 is divisible by 3.
• The same number is also divisible by 11.

Concept / Approach:
We apply the two divisibility rules:
Divisibility by 3: Sum of digits must be divisible by 3.Divisibility by 11: (Sum of digits in odd positions) − (sum of digits in even positions) must be a multiple of 11 (including 0).Positions are counted from the rightmost digit (units) as position 1.

Step-by-Step Solution:
Step 1: Use the rule for 3.Digits: 4, 7, 6, a, b, 0 → sum = 17 + a + b.For divisibility by 3, 17 + a + b must be a multiple of 3.Step 2: Use the rule for 11.Label positions from right: 0 (1), b (2), a (3), 6 (4), 7 (5), 4 (6).Odd positions: 0 + a + 7 = a + 7.Even positions: b + 6 + 4 = b + 10.Difference: (a + 7) − (b + 10) = a − b − 3.This difference must be 0, 11, −11, etc.Step 3: Solve for non-zero digits from the options.Check option (8, 5): a = 8, b = 5.For 11: a − b − 3 = 8 − 5 − 3 = 0 → multiple of 11 satisfied.For 3: digit sum = 4 + 7 + 6 + 8 + 5 + 0 = 30 → divisible by 3.

Verification / Alternative check:
Construct the number 476850. Check 476850 ÷ 3 = 158950 (integer). For 11, apply the 11-rule difference: (0 + 8 + 7) − (5 + 6 + 4) = 15 − 15 = 0, which is a multiple of 11. Hence 476850 is divisible by both 3 and 11.

Why Other Options Are Wrong:
Other pairs (6, 2), (8, 2), (6, 5) and (5, 2) either fail the 11-rule or do not satisfy the 3-rule when substituted as (a, b). Only (8, 5) satisfies both divisibility conditions simultaneously.

Common Pitfalls:
Students may mislabel positions for the 11-rule (starting from the left instead of the right), or forget that the difference can be zero. Some try to test divisibility by long division for each option, which is time consuming compared to using the rules directly.

Final Answer:
The hundreds and tens digits are 8 and 5 respectively.