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A realtor uses a lock box to store the keys to a house that is for sale. The access code for the lock box consists of four digits. The first digit cannot be 8 and the last digit must be even. How many different codes are available? (Note that 0 is considered an even number) The number of different codes available is (Type a whole number)

Ask by Sullivan Fowler. in the United States
Feb 25,2025

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Answer

There are 4,500 different codes available.

Solution

We have a four-digit code where: • The first digit can be any digit except 8 (so there are 9 choices: 0, 1, 2, 3, 4, 5, 6, 7, 9). • The second and third digits can be any digit from 0 to 9 (10 choices each). • The fourth digit must be even (choices are 0, 2, 4, 6, 8; hence, 5 choices). To get the total number of codes, multiply the possibilities together: 9 (first digit) × 10 (second digit) × 10 (third digit) × 5 (fourth digit) = 4500 Thus, there are 4500 different codes available.

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To determine how many different access codes are available, let's break it down step by step. 1. The first digit can be any digit from 0 to 9, except for 8. This gives us 9 possible choices (0, 1, 2, 3, 4, 5, 6, 7, 9). 2. The second digit has no restrictions, so there are 10 possible choices (0 through 9). 3. The third digit also has no restrictions, providing another 10 choices. 4. The last digit must be even. The possible even digits are 0, 2, 4, 6, and 8, which gives us 5 choices. Now, we calculate the total number of different codes by multiplying the choices for each digit: Total codes = (choices for first digit) × (choices for second digit) × (choices for third digit) × (choices for last digit) Total codes = 9 × 10 × 10 × 5 = 4500. So, the number of different codes available is 4500.

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