Question

A number \( x = 3.1212121212\ldots \). By what least number should we multiply \( x \), so that we obtain an integer?

• A. 3
• B. 9
• C. 30
• D. 33
• E. 99

Hint:

For repeating decimals, express the decimal part as a fraction to simplify the problem.

Answer 1

The Correct Option is D

Step 1: Express the number as a fraction.
Let \( x = 3.1212121212\ldots \), which is a repeating decimal. We can express this as: \[ x = 3 + 0.1212121212\ldots \] Let \( y = 0.1212121212\ldots \). Then we have: \[ y = 0.\overline{12}. \] Step 2: Convert the repeating decimal to a fraction.
To convert \( y = 0.\overline{12} \) to a fraction, multiply both sides by 100: \[ 100y = 12.\overline{12}. \] Now subtract \( y = 0.\overline{12} \) from \( 100y = 12.\overline{12} \): \[ 100y - y = 12.\overline{12} - 0.\overline{12}, \] \[ 99y = 12, \] \[ y = \frac{12}{99} = \frac{4}{33}. \] Step 3: Combine the integer and the fraction.
Thus, we have: \[ x = 3 + \frac{4}{33} = \frac{99}{33} + \frac{4}{33} = \frac{103}{33}. \] Step 4: Find the least multiplier to make \( x \) an integer.
To make \( x = \frac{103}{33} \) an integer, multiply both the numerator and denominator by 33: \[ x = \frac{103 \times 33}{33 \times 33} = \frac{3399}{1089}. \] Thus, multiplying by 33 will make \( x \) an integer.