Recover the missing tens digit in a product: 2013 × _1 = 62403. Identify the tens digit so that the multiplier becomes a two-digit number ending in 1.

Introduction / Context:
The multiplier is a two-digit number ending with 1. We are asked to find its tens digit. This is a reverse-multiplication problem where recognizing nearby multiples simplifies the task dramatically.

Given Data / Assumptions:

• Known multiplicand: 2013.
• Product: 62403.
• Multiplier format: _1 (a two-digit number whose ones digit is 1).

Concept / Approach:
Check 2013 × 31, since 31 is a common candidate (30 + 1). Multiplying by 30 then adding one extra 2013 is a quick mental method. If it matches the target product, the tens digit is 3; otherwise try 21, 41, etc. The structure strongly suggests 31 due to the product’s size relative to 2013 × 30.

Step-by-Step Solution:
Compute 2013 × 30 = 60390.Add one more 2013 (for ×31): 60390 + 2013 = 62403.Hence, multiplier = 31, so the tens digit is 3.

Verification / Alternative check:
Try 2013 × 21 = 2013 × (20 + 1) = 40260 + 2013 = 42273, which is too small. Try 41: 2013 × 41 = 82533, too large. Therefore 31 is uniquely correct.

Why Other Options Are Wrong:
Tens digits 1, 2, or 4 correspond to 11, 21, or 41, none of which produce 62403.

Common Pitfalls:
Misplacing zeros during the ×30 step; forgetting to add the extra 2013 for the +1 part; reading the question as asking for the whole multiplier rather than only the tens digit.

Final Answer:
3