All possible three digit numbers except one!

What is x x ?

1/x = 0. 000 001 002 003 004 005 .... 995 996 997 999 000 001 002 003 004 ....

and so on recurring.(that is all three digit numbers starting from 001 and continuing to the last digit 999 and then repeating. Wait.. there's a number missing in the pattern...does it have something to do with the answer?)

(And yes, there's no 998 between 997 and 999)

This is not a hint, but the answer may have something to do with the missing number! ... actually I don't know ... figure it out yourself. Well another hint: try out for 1/81. Hope you find a pattern!


The answer is 998001.

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1 solution

Rishabh Sood
Jun 20, 2017

Well.. this seems to be excellent.... for the fact that, 1/81 gives you 0.0123456790.., that is all possible single digit numbers except 8, and 1/9801 gives you 0.000102...996997999000.., that is all possible 2 digit numbers except 98. Similarly, 1/998001 gives you all possible 3 digit numbers except 998. So... you know what's the pattern? See..

  • 1 9 2 \frac{1}{9^2} => all possible 1 digit numbers except 8.
  • 1 9 9 2 \frac{1}{99^2} => all possible 2 digit numbers except 98.
  • 1 99 9 2 \frac{1}{999^2} => all possible 3 digit numbers except 998.
  • 1 999... n t i m e s 2 \frac{1}{999...n times^2} => all possible n digit numbers except 999..(n-1 times)8.

Actually if you want to create any decimal like=> 0.abcdefghijklmno, then the fraction for it will be : a b c d e f g h i j k l m n o 999999999999999 \frac{abcdefghijklmno}{999999999999999} , that is the digits after decimal divided by 9 repeated number of times as the number of digits!

Nice observation. Any logical reason behind this pattern?

Ashish Menon - 3 years, 11 months ago

It is not clear what the pattern is. If you had a \ldots in between 0006 and 9997, I would better understand (but still not perfectly) what you are trying to say.

Calvin Lin Staff - 3 years, 11 months ago

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