2017 AMC 8 Problem 20

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To help you master 2017 AMC 8 Problem 20, this page includes:

The original AMC problem to try, plus the correct answer

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2017 AMC8 Problem 20

Solution

Solution 1

Constructive Counting (Most Restricted First)

Build the four-digit number by filling in the digits one by one. To avoid complicated overlaps, we will select the digits with the most rules first: the units digit and the thousands digit.

By the Multiplication Principle, if we break a process down into independent steps, we can multiply the number of valid options at each step to find the total number of possible combinations.

Step 1: Calculate the total number of possible integers

The problem states the integer is chosen from 1000 to 9999 inclusive. We find the total number of integers in this range by subtracting the lower bound from the upper bound and adding 1.

Total integers = 9999 - 1000 + 1 = 9000

Step 2: Identify the restrictions for favorable outcomes

We are building a four-digit number using the 10 available digits from 0 to 9. The number must be odd, so the units digit is restricted to the odd digits. The number cannot start with zero, so the thousands digit cannot be 0. Furthermore, all four digits must be distinct.

Step 3: Count the options for the units and thousands digits

Always start by filling the positions with the most restrictions. The units digit must be odd. Once we choose the units digit, we move to the thousands digit. The thousands digit cannot be 0, and because the digits must be distinct, it cannot be the digit we just used for the units place. Since the odd units digit is definitely not 0, these are two entirely separate restrictions. We subtract 2 from our 10 original digits, leaving 8 choices for the thousands place.

Units choices = 5 (1, 3, 5, 7, 9)

Thousands choices = 10 - 1 (for 0) - 1 (for units digit) = 8

Step 4: Count the options for the remaining digits

Now we choose the hundreds digit. It can be any digit, including 0, except the two digits already used for the thousands and units places. Then we choose the tens digit, which can be anything except the three digits already used. We multiply all the choices together to find the number of favorable integers.

Hundreds choices = 10 - 2 = 8

Tens choices = 10 - 3 = 7

Favorable outcomes = 5 \times 8 \times 8 \times 7 = 2240

Step 5: Calculate the final probability

The probability is the number of favorable outcomes divided by the total number of possible integers. We simplify the resulting fraction by dividing the numerator and denominator by common factors (first by 10, then by 4).

Probability = \frac{2240}{9000}

\frac{2240}{9000} = \frac{224}{900} = \frac{56}{225}

Correct answer: B. $\frac{56}{225}$
The exact answer is

\frac{56}{225}

, which corresponds to choice B. We can verify the magnitude: the probability of an integer being odd is exactly

\frac{1}{2}

, and the probability of 4 digits being distinct is roughly

1 \times 0.9 \times 0.8 \times 0.7 \approx 0.5

. So we expect the answer to be roughly

0.5 \times 0.5 = 0.25

. Our fraction

\frac{56}{225}

is very close to

\frac{56}{224} = 0.25

, which gives us strong confidence in the result.

Solution 2

Constructive Counting by Cases

Instead of starting from the end, we can build the number starting from the thousands digit. Because the remaining choices for the odd units digit depend on whether the thousands digit is odd or even, we split the problem into two distinct cases.

When a single sequence of choices depends too heavily on previous selections, we use the Addition Principle to split the problem into non-overlapping cases, calculate the total for each case separately, and add the results together.

Step 1: Divide the problem into cases based on the first digit

If we choose the thousands digit first, we do not know whether it is odd or even, which changes how many choices are left for the odd units digit. To fix this, we split the problem into two mutually exclusive cases based on the parity of the thousands digit.

Step 2: Calculate the number of ways for Case 1 (thousands digit is odd)

If the thousands digit is odd, there are 5 possible choices (1, 3, 5, 7, 9). We then jump to the units digit. The units digit must also be odd and distinct, so one odd digit has been consumed, leaving 4 choices. The hundreds digit can be any of the 8 remaining unassigned digits, and the tens digit can be any of the 7 remaining digits.

Thousands choices (odd) = 5

Units choices (odd) = 4

Hundreds choices = 8

Tens choices = 7

Case 1 total = 5 \times 4 \times 8 \times 7 = 1120

Step 3: Calculate the number of ways for Case 2 (thousands digit is even)

If the thousands digit is even and non-zero, there are 4 choices (2, 4, 6, 8). We then jump to the units digit. It must be odd, and since the thousands digit is even, none of the 5 odd digits have been used yet, leaving 5 choices. The hundreds and tens digits again have 8 and 7 choices, respectively.

Thousands choices (even, non-zero) = 4

Units choices (odd) = 5

Hundreds choices = 8

Tens choices = 7

Case 2 total = 4 \times 5 \times 8 \times 7 = 1120

Step 4: Combine the cases and compute the probability

Since these cases do not overlap and cover all possibilities for a valid starting digit, we add them to get the total number of favorable integers. We then divide by 9000 to find the probability.

Favorable outcomes = 1120 + 1120 = 2240

Probability = \frac{2240}{9000} = \frac{56}{225}

Correct answer: B. $\frac{56}{225}$
The probability is

\frac{56}{225}

, matching choice B. By arriving at the exact same numerator

2240

using a different logic path and starting digit, we confirm that our accounting of the overlapping zero and odd constraints is perfectly accurate.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Definition of Probability:

P (A) = \frac{Number of favorable outcomes}{Total number of possible outcomes}

A is the specific event we want to occur. This formula applies when all outcomes in the sample space are equally likely. Used at the very end to turn our integer counts into a formal probability fraction.

Multiplication Principle (Rule of Product):

N = n_{1} \times n_{2} \times \dots \times n_{k}

n_i is the number of valid options for the i-th step. This applies when decisions are made in sequence and the number of available options at each step does not depend on the specific choices made in previous steps. Used to find the number of valid outcomes by multiplying the available choices for the thousands, hundreds, tens, and units digits.

Addition Principle (Rule of Sum):

N = N_{1} + N_{2} + \dots + N_{k}

N_i is the number of outcomes in the i-th case. This applies only when the cases are mutually exclusive (they cannot happen at the same time) and collectively exhaustive (they cover all possibilities). Used in Method 2 to safely combine the totals from the odd thousands digit case and the even thousands digit case.

Key Takeaway: Remember This Strategy for Next Time

A combinatorics problem asks you to arrange items (like digits) where some positions have strict rules (such as not being 0, or being an odd number).

Try: Fill the most restricted positions first.

Why: By dealing with the strict rules early, you ensure that you don't accidentally run out of valid options. The remaining, less restricted positions usually just require counting what is left over.

Making a choice in one position changes the number of available options for a future position (e.g., picking an odd first digit leaves fewer odd digits for the last place).

Try: Break the problem into mutually exclusive cases.

Why: Separating the scenarios mathematically untangles the overlapping rules. Within each case, the number of choices remains constant and can be safely multiplied without ambiguity.

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2017 AMC 8 Problem 20

In This Page:

2017 AMC8 Problem 20