Question

\(\large \begin{align} \color{black}{
0.abababab\cdot \cdot \cdot \cdot \cdot \cdot \cdot \times n \hspace{.33em}\\~\\
\normalsize \text{is an integer.} \hspace{.33em}\\~\\
\normalsize \text{a and b are single digit natural number.} \hspace{.33em}\\~\\
\normalsize \text{find the greatest 3 digit value that 'n' can take.} \hspace{.33em}\\~\\
}\end{align}\)

Answer 1

990 ?

Answer 2

yes , but m not sure how that came

Answer 3

that just came haha

Answer 4

(jk)

Answer 5

\(\large \begin{align} \color{black}{
0.abababab\cdot \cdot \cdot \cdot \cdot \cdot \cdot \times n \hspace{.33em}\\~\\
=\dfrac{ab}{99} \times n \hspace{.33em}\\~\\
}\end{align}\)

how to proceed

Answer 6

for that stuff to be an integer for arbitrary \(a, b\) ,
you need \(n\) to be a multiple of \(99\) ?

Answer 7

yes

Answer 8

i m still in doubt ,what this line means exactly

\(\normalsize \text{find the greatest 3 digit value that 'n' can take.} \hspace{.33em}\\~\\\)

Answer 9

999 is the greatest 3 digit number, yes ?

Answer 10

yes

Answer 11

we have \(n = 99k\) for some integer \(k\)
\(k = 2-10\) gives you 3 digit numbers, out of which 99*10 is the greatest one

Answer 12

ok and if the question was


\(\normalsize \text{find the leat 3 digit value that 'n' can take.} \hspace{.33em}\\~\\\)

then it would be \(99\times 2\)

Answer 13

thats true only when \(\gcd(99, (ab)_{10}) =1\) :

\[\dfrac{ab}{99}\times n\]

Answer 14

example :
\[\dfrac{18}{99}\times n\]
least 3 digit value for \(n\) is not \(99*2\) anymore..

Answer 15

so \(ab_{10}\) should be atleast \(50\)
for this

\(\normalsize \text{ greatest 3 digit value that 'n' can take.} \hspace{.33em}\\~\\\)

Answer 16

if we take \(n=2*99\)