Divisibility by 11 (alternating-sum test) Find the least digit * so that 97215*6 is divisible by 11.

Introduction / Context:
A number is divisible by 11 if the alternating-sum of its digits is a multiple of 11 (including 0). This problem asks for the smallest digit * that makes 97215*6 divisible by 11.

Given Data / Assumptions:

• Number pattern: 9 7 2 1 5 * 6.
• Alternating sum taken left to right with +, −, +, −, …
• Result must be 0, ±11, ±22, …

Concept / Approach:
Compute S = 9 − 7 + 2 − 1 + 5 − * + 6. Set S equal to a multiple of 11 and solve for * among digits 0–9. Choose the minimal valid digit.

Step-by-Step Solution:
Compute without *: 9−7=2; +2=4; −1=3; +5=8; +6=14.Expression becomes 14 − *.Set 14 − * ≡ 0 (mod 11) ⇒ 14 − * = 11 (since 14 and * are digits-sized apart).Hence * = 3 (least valid digit).

Verification / Alternative check:
With * = 3, S = 14 − 3 = 11 → divisible by 11, so the number is divisible by 11.

Why Other Options Are Wrong:
1, 2, 5 give S = 13, 12, 9 respectively, none of which are multiples of 11.

Common Pitfalls:
Reversing the ± pattern; forcing S = 0 only, when any multiple of 11 suffices; choosing the largest valid digit instead of the least.

Final Answer:
3