476xy0 is divisible by both 3 and 11. The nonzero digits in the hundreds and tens places (x, y) are respectively:

Given data

• Number format: 4 7 6 x y 0 (six digits).
• Divisible by 3 and by 11; x and y are nonzero digits.

Concept / Approach

• Divisibility by 3: Sum of digits multiple of 3.
• Divisibility by 11: (sum of digits in odd positions) − (sum of digits in even positions) ≡ 0 (mod 11).

Step-by-step constraints

Sum for /3: 4 + 7 + 6 + x + y + 0 = 17 + x + y ⇒ need x + y ≡ 1 (mod 3).For /11: (4 + 6 + y) − (7 + x + 0) = 3 + (y − x) ≡ 0 (mod 11) ⇒ y − x ≡ 8 (mod 11).Feasible small solutions for digits (1–9): y − x = 8 or y − x = −3 (since −3 ≡ 8 mod 11).Check candidates:• x=1, y=9 ⇒ x + y = 10 ≡ 1 (mod 3) ✔• x=5, y=2 ⇒ y − x = −3 and x + y = 7 ≡ 1 (mod 3) ✔• x=8, y=5 ⇒ y − x = −3 and x + y = 13 ≡ 1 (mod 3) ✔

Verification

All three pairs (1,9), (5,2), (8,5) satisfy both divisibility rules. Therefore, each yields a number divisible by 3 and 11.

Common pitfalls

• Forgetting that the 11-rule allows differences of ±11k, not only 0.

Final Answer

Valid ordered pairs (x, y): (1, 9), (5, 2), (8, 5).