Question

Let D be recurring decimal of the form, \( D = 0.a_1a_2a_1a_2a_1a_2\ldots \), where digits \( a_1 \) and \( a_2 \) lie between 0 and 9. Further, at most one of them is zero. Then which of the following numbers necessarily produces an integer, when multiplied by D?

• A. 18
• B. 108
• C. 198
• D. 288

Hint:

Write the repeating decimal as a fraction with denominator 99, then ask which number is guaranteed to cancel that denominator for every allowed digit pair.

Answer 1

The Correct Option is C

Step 1: Convert D into a fraction:
\( D = 0.\overline{a_1a_2} = \dfrac{10a_1 + a_2}{99} \).

Step 2: What must the multiplier do?
The multiplier \( m \) must make \( m\cdot\dfrac{10a_1+a_2}{99} \) an integer for every allowed pair of digits. Since \( 10a_1 + a_2 \) can be as small as 1 (take \( a_1 = 0,\ a_2 = 1 \); only one digit is zero), \( D = \dfrac{1}{99} \) is allowed. Then \( m \) must itself be a multiple of 99.

Step 3: Check the options:
\( 99 = 9 \times 11 \). 18, 108 and 288 are not multiples of 11, so they are not multiples of 99. \( 198 = 2 \times 99 \) is a multiple of 99.

Step 4: Verify:
For any digits, \( 198 \cdot D = 2(10a_1+a_2) \), an integer.

Final Answer:
The correct option is (3): 198