The first four digits of the six-digit initial password for a shopper's card at a certain grocery store is the...

Understanding the Question

We need to find a customer's birth date (month and day) given that their password ends in "16".

Given Information

• Password format: \(\mathrm{DDMMXY}\) where:
  • First 4 digits (DDMM) = birthday in day-month format
  • 5th digit (X) = units digit of \(\mathrm{7 \times (1st\,digit + 3rd\,digit)}\)
  • 6th digit (Y) = units digit of \(\mathrm{3 \times (2nd\,digit + 4th\,digit)}\)
• Known constraint: The password ends in "16", so X = 1 and Y = 6

What We Need to Determine

We need to find the unique values for DD (day) and MM (month).

For sufficiency: We need exactly one valid date that produces a password ending in "16".

Key Insights

Since the password ends in "16":

• The 5th digit = 1 → This means \(\mathrm{7 \times (1st\,digit + 3rd\,digit)}\) must end in 1
  • This only happens when \(\mathrm{(1st + 3rd) = 3\,or\,13}\)
• The 6th digit = 6 → This means \(\mathrm{3 \times (2nd\,digit + 4th\,digit)}\) must end in 6
  • This only happens when \(\mathrm{(2nd + 4th) = 2, 12,\,or\,22}\)

These constraints significantly limit our possible dates.

Analyzing Statement 1

Statement 1 tells us: The password begins with "21" and its fourth digit is 1.

This gives us the partial date: 21_1

What Statement 1 Tells Us

• Day = 21 (so 1st digit = 2, 2nd digit = 1)
• Month ends with 1 (4th digit = 1)
• We need to find the 3rd digit (first digit of month)

Finding the Missing Digit

Since the 5th digit must be 1, we need (1st digit + 3rd digit) to equal either 3 or 13.

• 1st digit = 2
• So: \(\mathrm{2 + 3rd\,digit = 3\,or\,13}\)
• Therefore: 3rd digit = 1 or 11

Since digits must be single values (0-9), the 3rd digit must equal 1.

This gives us the complete date: 21/01 (January 21st)

Quick Verification

Let's verify this produces a password ending in 16:

• 5th digit: \(\mathrm{7 \times (2 + 1) = 7 \times 3 = 21}\) → units digit = 1 ✓
• 6th digit: \(\mathrm{3 \times (1 + 1) = 3 \times 2 = 6}\) → units digit = 6 ✓

Statement 1 gives us exactly one valid date.

[STOP - Statement 1 is Sufficient!]

This eliminates choices B, C, and E.

Analyzing Statement 2

Important: We now forget Statement 1 completely and analyze Statement 2 independently.

Statement 2 tells us: The sum of the first and third digits is 3, and the second digit is 1.

This gives us:

• \(\mathrm{(1st\,digit + 3rd\,digit) = 3}\)
• 2nd digit = 1
• Date format: _1__

What We Need to Find

We need to determine which combination of 1st and 3rd digits (that sum to 3) creates a valid date.

Finding All Valid Dates

With 2nd digit = 1, possible days are: 01, 11, 21, 31

For (1st + 3rd) = 3, let's check each possibility:

• If 1st = 0, 3rd = 3 → Date: 01/03 (March 1st) ✓
• If 1st = 1, 3rd = 2 → Date: 11/02 (February 11th) ✓
• If 1st = 2, 3rd = 1 → Date: 21/01 (January 21st) ✓
• If 1st = 3, 3rd = 0 → Date: 31/00 (invalid - no month 00) ✗

We have three valid dates that satisfy Statement 2's conditions. Since we cannot determine a unique date, Statement 2 alone is NOT sufficient.

This eliminates choices B and D.

The Answer: A

Statement 1 alone gives us exactly one date (January 21st), while Statement 2 alone gives us three possible dates.

Answer Choice A: "Statement 1 alone is sufficient, but Statement 2 alone is not sufficient."