Kaushik786 wrote:
If x lies between zero and one, is the tenths digit of x zero ?
1. 99x is an integer
2. 81x is an integer
1. 99x is an integer
So x can be \(\frac{a}{99}\), where a is an integer from 1 to 98, That is x is any of \(\frac{1}{99}, \ \\frac{2}{99}, \ \\frac{3}{99},...... \ \\frac{98}{99}, \ \\)
If x is \(\frac{1}{99}, \ \till \ \\frac{9}{99}\), the tenth digit is 0 => \(\frac{1}{99}=0.010101...\)
If x is \(\frac{10}{99}\), the tenth digit is 1, as \(\frac{10*1}{99}=10*0.010101...=0.10101...\)
Insufficient
2. 81x is an integer
So x can be \(\frac{a}{81}\), where a is an integer from 1 to 80, That is x is any of \(\frac{1}{81}, \ \\frac{2}{81}, \ \\frac{3}{81},...... \ \\frac{80}{81}, \ \\)
If x is \(\frac{1}{81}, \ \till \ \\frac{8}{81}\), the tenth digit is 0 => \(\frac{1}{81}=0.012345...\)
If x is \(\frac{27}{81}=\frac{1}{3}\), the tenth digit is 3, as \(\frac{1}{3}=0.3333...\)
Insufficient
Combined
x has to have a denominator equal to the LCM(99,81) or 891.
Also, the maximum value of numerator has to be just less than 99, as restricted by statement I.
Then \(x<\frac{99}{891} \ \ or \ \ x<\frac{1}{11}.....x<0.090909..\)
Here the tenths digit will always be 0.
I always learn from your robust solution but this one is confusing a little bit. I don't know if it's because of the formatting not appearing well or my misreading it.
Pls check out my solution, see how i got C and help me point out the error in my process bcos i don't understand how i concluded differently that the tenth digit of x must always be non zero while you got that it must always be zero.