Question & Answer
QUESTION

\[xy\] is a number that is divided by \[ab\] where \[xy < ab\] and gives a result $0.xyxyxy...$. Then what is the value of $ab$?
(a) 11
(b) 33
(c) 99
(d) 66

ANSWER Verified Verified
Hint: We will first write $0.xyxyxy...$ in its expanded form and then take factor $xy$ common out of the expanded form and solve the other factor to find the value of $ab$.

Complete step-by-step answer:

The number $xy$ with $x$ at tens place and $y$ at ones place Can be written as: $10x+y$.

Now, multiplying 100 in numerator and denominator, we can write $0.xy$ as below:

$\begin{align}

  & 0.xy=\dfrac{0.xy\times 100}{100} \\

 & =\dfrac{xy}{100} \\

\end{align}$

Again, $0.xyxy$ can be expanded as,

$0.xyxy=0.xy+0.00xy$

Multiplying 10000 in numerator and denominator of $0.00xy$, we get

$\begin{align}

  & 0.xyxy=\dfrac{xy}{100}+\dfrac{0.00xy\times 10000}{10000} \\

 & =\dfrac{xy}{100}+\dfrac{xy}{10000} \\

\end{align}$

Similarly, we can write $0.xyxyxy...$ as below,

$\begin{align}

  & 0.xyxyxy...=0.xy+0.00xy+0.0000xy+... \\

 & =\dfrac{xy}{100}+\dfrac{xy}{10000}+\dfrac{0.0000xy\times 1000000}{1000000}+... \\

 & =\dfrac{xy}{100}+\dfrac{xy}{10000}+\dfrac{xy}{1000000}+... \\

\end{align}$

Taking $\dfrac{xy}{100}$ common, we get,

$=\dfrac{xy}{100}\left( 1+\dfrac{1}{100}+\dfrac{1}{10000}+... \right)$

Here, the terms of the second factor $1,\dfrac{1}{100},\dfrac{1}{10000},\,...$ are in a geometric progression.

Also, we know that for a geometric progression with first term $a$ and common ratio $r$, such that, $0 < r < 1$, sum say $S$, of its infinite series is given by,

$S=\dfrac{a}{1-r}\cdots \cdots \left( i \right)$

Here, in geometric progression $1,\dfrac{1}{100},\dfrac{1}{10000},\,...$, $a=1$ and

$r=\dfrac{1}{100}$.

Therefore, using equation $\left( i \right)$, its sum is given as,

$\begin{align}

  & 1+\dfrac{1}{100}+\dfrac{1}{10000}+... \\

 & =\dfrac{1}{1-\dfrac{1}{100}} \\

\end{align}$

Taking LCM in denominator, we get,

$=\dfrac{1}{\dfrac{1-100}{100}}$

Writing 100 in numerator, we get,

$\begin{align}

  & \dfrac{100}{1-100} \\

 & =\dfrac{100}{99} \\

\end{align}$

Therefore, $0.xyxyxy...$ can be written as,

$\begin{align}

  & 0.xyxyxy...=\dfrac{xy}{100}\times \dfrac{100}{99} \\

 & =\dfrac{xy}{99} \\

\end{align}$

Hence, $xy$ is divided by 99 to get $0.xyxyxy...$.

Therefore, the correct answer is option (c).

Note: In this type of question, do not confuse $xy$ to be the product of variables $x$ and $y$. It is not a product but the digits of the number in tens place and ones place.