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Sample Question 20:
Step 1: Interpret that the repeating decimal \(0.\overline{ab}\) is equal to \(\frac{10a+b}{99}\) as each successive term has more decimal places.
Step 2: When expressed in its simplest form, the denominator of the fraction will always be a factor of the number 99, which can be factored into 3, 3, 11. This gives the possible factors as 1, 3, 9, 11, 33, and 99.
Step 3: However, since the problem specifies that \(a\) and \(b\) are not both nine and not both zero, the denominator 1 cannot be achieved because it would require the numerator to be 99. This leaves us with 5 potential denominators: 3, 9, 11, 33, and 99.
The final answer is \(5\), which corresponds to choice \(\mathrm{(C)}\).
Suppose that \(a\) and \(b\) are digits, not both nine and not both zero, and the repeating decimal \(0.\overline{ab}\) is expressed as a fraction in lowest terms. How many different denominators are possible?
\(\text{(A) }3 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }8 \qquad \text{(E) }9\)
Answer Keys
Question 20: CSolutions
Question 20Step 1: Interpret that the repeating decimal \(0.\overline{ab}\) is equal to \(\frac{10a+b}{99}\) as each successive term has more decimal places.
Step 2: When expressed in its simplest form, the denominator of the fraction will always be a factor of the number 99, which can be factored into 3, 3, 11. This gives the possible factors as 1, 3, 9, 11, 33, and 99.
Step 3: However, since the problem specifies that \(a\) and \(b\) are not both nine and not both zero, the denominator 1 cannot be achieved because it would require the numerator to be 99. This leaves us with 5 potential denominators: 3, 9, 11, 33, and 99.
The final answer is \(5\), which corresponds to choice \(\mathrm{(C)}\).