What is the probability that a randomly selected three-digit number has the property that one digit is equal to the product of the other two? Express your answer as a common fraction.
13/225
To find the probability that a randomly selected three-digit number has the property that one digit is equal to the product of the other two, we can break down the problem into cases.
Case 1: The hundreds digit is equal to the product of the tens and units digit.
In this case, there are 9 possible options for the hundreds digit (1 to 9), and for each hundreds digit, there are 9 options for the tens digit (0 to 9, excluding the hundreds digit itself), and 10 options for the units digit (0 to 9). This gives us a total of 9 * 9 * 10 = 810 possibilities.
Case 2: The tens digit is equal to the product of the hundreds and units digit.
In this case, there are 10 possible options for the tens digit (0 to 9), and for each tens digit, there are 9 options for the hundreds digit (1 to 9), and 9 options for the units digit (1 to 9, excluding the tens digit itself). This gives us a total of 10 * 9 * 9 = 810 possibilities.
Case 3: The units digit is equal to the product of the hundreds and tens digit.
In this case, there are 10 possible options for the units digit (0 to 9), and for each units digit, there are 9 options for the hundreds digit (1 to 9), and 10 options for the tens digit (0 to 9). This gives us a total of 10 * 9 * 10 = 900 possibilities.
Therefore, the total number of three-digit numbers that have the desired property is 810 + 810 + 900 = 2520.
Since there are 9000 three-digit numbers in total (from 100 to 999), the probability is given by:
Probability = (desired outcomes) / (total outcomes) = 2520 / 9000 = 7/25.
So, the probability that a randomly selected three-digit number has the property that one digit is equal to the product of the other two is 7/25.
To solve this problem, we need to consider all possible cases and count the favorable outcomes.
Let's break down the problem into cases:
Case 1: One of the digits is zero.
If one of the digits is zero, then the other two digits must also be zero since any product involving zero will result in zero. Therefore, there are no favorable outcomes in this case.
Case 2: Two of the digits are equal to each other.
Let's consider the possible pairs of two equal digits: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), and (9, 9).
For each pair, we have one remaining digit that can be multiplied with the equal digits. So for each pair, there are 9 possible third digits.
Therefore, the total number of favorable outcomes in this case is 9 * 9 = 81.
Case 3: All three digits are distinct.
In this case, we have six possible arrangements for the three digits: ABC, ACB, BAC, BCA, CAB, and CBA.
For each arrangement, there is only one possible product relationship, such as A = B * C. We need to find the number of sets of three digits where one digit is equal to the product of the other two.
Let's consider the distinct digits in increasing order: A, B, and C. We need to find the number of sets that satisfy A = B * C.
For a specific value of A, there are only two possible values for B and C (since they must be distinct), such that A = B * C. Without loss of generality, let's assume A < B < C.
For A = 1, there are no possible values for B and C since no two digits multiply to give one.
For A = 2, there is only one possible set: B = 1 and C = 2.
For A = 3, there is only one possible set: B = 1 and C = 3.
For A = 4, there is only one possible set: B = 1 and C = 4.
For A = 5, there are two possible sets: B = 1 and C = 5 or B = 2 and C = 5.
For A = 6, there are two possible sets: B = 1 and C = 6 or B = 2 and C = 6.
For A = 7, there are three possible sets: B = 1 and C = 7, B = 2 and C = 7, or B = 3 and C = 7.
For A = 8, there are three possible sets: B = 1 and C = 8, B = 2 and C = 8, or B = 4 and C = 8.
For A = 9, there are four possible sets: B = 1 and C = 9, B = 2 and C = 9, B = 3 and C = 9, or B = 4 and C = 9.
Therefore, the total number of favorable outcomes in this case is 1 + 1 + 1 + 2 + 2 + 3 + 3 + 4 = 17.
Now, let's calculate the probability by dividing the total number of favorable outcomes (81 + 17 = 98) by the total number of possible three-digit numbers, which is 900 (9 choices for the hundreds digit, 10 choices for the tens digit, and 10 choices for the units digit).
Thus, the probability is 98/900, which simplifies to 7/64.
Therefore, the probability that a randomly selected three-digit number has the property that one digit is equal to the product of the other two is 7/64.