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31 Dec 2023, 10:28
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Peter has two fair six-sided dice. One is labeled with the first six prime numbers, and the other with their reciprocals. If he rolls both dice once, what is the probability that the product of the resulting numbers is NOT a terminating decimal? (A terminating decimal is any decimal that has only a finite number of nonzero digits. For example, 24, 0.82, and 5.096 are three terminating decimals.)
A. \(\frac{1}{36}\)
B. \(\frac{1}{6}\)
C. \(\frac{5}{9}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)
A. \(\frac{1}{36}\)
B. \(\frac{1}{6}\)
C. \(\frac{5}{9}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)
31 Dec 2023, 10:28
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Official Solution:
Peter has two fair six-sided dice. One is labeled with the first six prime numbers, and the other with their reciprocals. If he rolls both dice once, what is the probability that the product of the resulting numbers is NOT a terminating decimal? (A terminating decimal is any decimal that has only a finite number of nonzero digits. For example, 24, 0.82, and 5.096 are three terminating decimals.)
A. \(\frac{1}{36}\)
B. \(\frac{1}{6}\)
C. \(\frac{5}{9}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)
A reduced fraction \(\frac{a}{b}\) (meaning that the fraction is already in its simplest form, so reduced to its lowest term) can be expressed as a terminating decimal if and only if the denominator \(b\) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as the denominator \(250\) equals \(2*5^3\). The fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and the denominator \(10=2*5\).
Note that if the denominator already consists of only 2s and/or 5s, then it doesn't matter whether the fraction is reduced or not.
For example, \(\frac{x}{2^n5^m}\), (where \(x\), \(n\), and \(m\) are integers) will always be a terminating decimal.
(We need to reduce the fraction in case the denominator has a prime other than 2 or 5, to see whether it can be reduced. For example, the fraction \(\frac{6}{15}\) has 3 as a prime in the denominator, and we need to know if it can be reduced.)
Thus, for the product of the numbers not to be a terminating decimal, its denominator, after reduction, must have primes other than 2 and 5.
We have two dice, one is labeled with the first six prime numbers: 2, 3, 5, 7, 11, and 13, and the other with their reciprocals: \(\frac{1}{2}, \ \frac{1}{3}, \ \frac{1}{5}, \ \frac{1}{7}, \ \frac{1}{11}\), and \(\frac{1}{13}\).
If the die with the reciprocals shows \(\frac{1}{2}\), or \(\frac{1}{5}\), then the product will always be a terminating decimal, regardless of the result on the second die.
• If the die with the reciprocals shows \(\frac{1}{3}\), then the die with primes should show 2, 5, 7, 11, or 13, for the product not to be a terminating decimal. The probability of that is \(\frac{1}{6}*\frac{5}{6}\).
• If the die with the reciprocals shows \(\frac{1}{7}\), then the die with primes should show 2, 3, 5, 11, or 13, for the product not to be a terminating decimal. The probability of that is \(\frac{1}{6}*\frac{5}{6}\).
• If the die with the reciprocals shows \(\frac{1}{11}\), then the die with primes should show 2, 3, 5, 7, or 13, for the product not to be a terminating decimal. The probability of that is \(\frac{1}{6}*\frac{5}{6}\).
• If the die with the reciprocals shows \(\frac{1}{13}\), then the die with primes should show 2, 3, 5, 7, or 11, for the product not to be a terminating decimal. The probability of that is \(\frac{1}{6}*\frac{5}{6}\).
Therefore, the overall probability is \(4*\frac{1}{6}*\frac{5}{6}=\frac{5}{9}\).
Answer: C
Peter has two fair six-sided dice. One is labeled with the first six prime numbers, and the other with their reciprocals. If he rolls both dice once, what is the probability that the product of the resulting numbers is NOT a terminating decimal? (A terminating decimal is any decimal that has only a finite number of nonzero digits. For example, 24, 0.82, and 5.096 are three terminating decimals.)
A. \(\frac{1}{36}\)
B. \(\frac{1}{6}\)
C. \(\frac{5}{9}\)
D. \(\frac{2}{3}\)
E. \(\frac{5}{6}\)
A reduced fraction \(\frac{a}{b}\) (meaning that the fraction is already in its simplest form, so reduced to its lowest term) can be expressed as a terminating decimal if and only if the denominator \(b\) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as the denominator \(250\) equals \(2*5^3\). The fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and the denominator \(10=2*5\).
Note that if the denominator already consists of only 2s and/or 5s, then it doesn't matter whether the fraction is reduced or not.
For example, \(\frac{x}{2^n5^m}\), (where \(x\), \(n\), and \(m\) are integers) will always be a terminating decimal.
(We need to reduce the fraction in case the denominator has a prime other than 2 or 5, to see whether it can be reduced. For example, the fraction \(\frac{6}{15}\) has 3 as a prime in the denominator, and we need to know if it can be reduced.)
Thus, for the product of the numbers not to be a terminating decimal, its denominator, after reduction, must have primes other than 2 and 5.
We have two dice, one is labeled with the first six prime numbers: 2, 3, 5, 7, 11, and 13, and the other with their reciprocals: \(\frac{1}{2}, \ \frac{1}{3}, \ \frac{1}{5}, \ \frac{1}{7}, \ \frac{1}{11}\), and \(\frac{1}{13}\).
If the die with the reciprocals shows \(\frac{1}{2}\), or \(\frac{1}{5}\), then the product will always be a terminating decimal, regardless of the result on the second die.
• If the die with the reciprocals shows \(\frac{1}{3}\), then the die with primes should show 2, 5, 7, 11, or 13, for the product not to be a terminating decimal. The probability of that is \(\frac{1}{6}*\frac{5}{6}\).
• If the die with the reciprocals shows \(\frac{1}{7}\), then the die with primes should show 2, 3, 5, 11, or 13, for the product not to be a terminating decimal. The probability of that is \(\frac{1}{6}*\frac{5}{6}\).
• If the die with the reciprocals shows \(\frac{1}{11}\), then the die with primes should show 2, 3, 5, 7, or 13, for the product not to be a terminating decimal. The probability of that is \(\frac{1}{6}*\frac{5}{6}\).
• If the die with the reciprocals shows \(\frac{1}{13}\), then the die with primes should show 2, 3, 5, 7, or 11, for the product not to be a terminating decimal. The probability of that is \(\frac{1}{6}*\frac{5}{6}\).
Therefore, the overall probability is \(4*\frac{1}{6}*\frac{5}{6}=\frac{5}{9}\).
Answer: C
14 Jul 2024, 14:25
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If the die with the reciprocals shows 1/ 2 or 1/5
then the product will always be a terminating decimal, regardless of the result on the second die.
Does this mean if the power of either 5 or 2 is 0, then also it would result in terminating decimal?
then the product will always be a terminating decimal, regardless of the result on the second die.
Does this mean if the power of either 5 or 2 is 0, then also it would result in terminating decimal?
