Problem Solving (PS) | The repeating decimal 1.ab, where a and b are different digits, is equ

N =\(1.\overline{ab}\)

We can write N as

N =1+\(\frac{ab}{99}\)

N=\( \frac{ 99 +(ab)}{99}\)

As the GCD of (99 + ab) & 99 =1, we should not have factor in common between ab and 99. Its given that ab are different numbers and we need the greatest possible value of ab, so ab can be98.

\( N =\frac{(99+98)}{99}\)

\( N =\frac{ 197 }{99}\)

n = 197

d = 99

197 + 99 =296

OptionA
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Statistics : Posted by gmatophobia • on 25 Nov 2022, 06:00 • Replies 1 • Views 64

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