Divisibility by 11 — If the 6-digit number 1x5x01 is divisible by 11, find the digit x that satisfies the rule for alternately signed sums.

Introduction / Context:Divisibility by 11 has a distinctive rule using alternating sums of digits. Applying that rule lets us solve for an unknown digit without long division.

Given Data / Assumptions:

• Number shape: 1 x 5 x 0 1 (six digits).
• Unknown: digit x (0 to 9).
• Requirement: number divisible by 11.

Concept / Approach:Rule: For a number d1 d2 d3 d4 d5 d6, compute S = (d1 + d3 + d5) - (d2 + d4 + d6). If S is 0 or a multiple of 11 (e.g., ±11, ±22), the number is divisible by 11.

Step-by-Step Solution:Assign digits: d1=1, d2=x, d3=5, d4=x, d5=0, d6=1.Compute odd-position sum: d1 + d3 + d5 = 1 + 5 + 0 = 6.Compute even-position sum: d2 + d4 + d6 = x + x + 1 = 2x + 1.Form S: 6 - (2x + 1) = 5 - 2x.Set S equal to 0 or ±11, ±22. Only 5 - 2x = -11 gives a valid digit: 2x = 16 → x = 8.

Verification / Alternative check:Substitute x=8 to get 185801. Alternating-sum S = (1+5+0) - (8+8+1) = 6 - 17 = -11, which is a multiple of 11, confirming divisibility.

Why Other Options Are Wrong:

• 6 / 4 / 9 / 2: These values do not make S equal to 0, ±11, or ±22; hence the number would not be divisible by 11.

Common Pitfalls:Mis-ordering positions for the alternating sum; forgetting that negative multiples like -11 also qualify; restricting checks only to S=0 and missing ±11 solutions.

Final Answer:8