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Questions 1-5 A company employee generates a series of five-digit product codes in accordance with the following rules: The codes use the digits 0, 1, 2, 3, and 4, and no other. Each digit occurs exactly once in any code. The second digit has a value exactly twice that of the first digit. The value of the third digit is less than the value of the fifth digit.

STEP 1: Symbolize the rules. Numbering the rules is a good way to stay organized. The numbers used below don’t match the indented rules (because the first rule only gives us the variables), but matching isn’t necessary. When the explanation refers to “rule number X”, it is using the numbers below. 0, 1, 2, 3, 4 1) Each ex. once 2) ___ ___ 1

2

2X 3) 3 < 5 STEP 2: Create the master set-up. Rule 2 looks like it will have the greatest effect on the game. Let’s start there. As it turns out, Rule 2 creates two scenarios: In Scenario # 1, you must have 1 and 2 in the first and second slots; in Scenario # 2, you must have 2 and 4 in the first and second slots. No other variables work. Scenario # 1

Scenario # 2

1

2

2

4

Now we can look at Rule 3. Rule 3 states that the third digit must be less than the fifth digit. This means two things: • •

The third slot can’t be the largest number remaining (because the fifth digit must be larger); The fifth slot can’t be the smallest number remaining (because the third digit must be smaller).

Let’s apply Rule 3 to each scenario separately. In scenario # 1, the digits 1 and 2 have been used. 0, 3, and 4 remain. Since the third slot cannot be the largest number remaining, it cannot be 4. It must be either 0 or 3. Since the fifth slot cannot be the smallest number remaining, it cannot be 0. It must be either 3 or 4. Scenario # 1

1

2

0/3

3/4

4

0

.

In scenario # 2, the digits 2 and 4 have been used. 0, 1, and 3 remain. Since the third slot cannot be the largest number remaining, it cannot be 3. It must be either 0 or 1. Since the fifth slot cannot be the smallest number remaining, it cannot be 0. It must be either 1 or 3. Scenario # 2

2

4

0/1

1/3

3

0

.

Your final set-up should look like this: Scenario # 1

Scenario # 2

1

2

2

4

0/3

3/4

4

0

0/1

1/3

3

0

1. If the last digit of an acceptable product code is 1, it must be true that the

.

.

(A) (B) (C) (D) (E)

first digit is 2 second digit is 0 third digit is 3 fourth digit is 4 fourth digit is 0

STEP 1: Classify the question. This is a local question because it gives us an additional condition – that the last digit of a code is 1. STEP 2: Fill in the local condition. Because the question tells us the last digit of the product code is 1, we must be in scenario # 2. This is because scenario # 1 requires that 1 be placed in the first slot. The first step is to fill in the local condition, along with what we already know, and see if we can make any further deductions. Scenario # 2

2

4

1

STEP 3: Make deductions. Since we know that the third slot can only be 0 or 1 in Scenario # 2, and 1 is already in the fifth slot, the third slot must be 0. Scenario # 2

2

4

0

1

3 is the only variable remaining. We can place it in the fourth slot. Scenario # 2

2

4

0

3

1

THE CORRECT ANSWER IS (A). THE FIRST DIGIT MUST BE 2.