Prove: If x is a positive real number, then somewhere in the infinite sequence {x, 2x, 3x, ...} there is a number containing the digit 7.
If x is a positive real number, then somewhere in the finite sequence {x, 2x, 3x, ..., nx} there is a number containing the digit 7. Find the minimum value of n.
Note: Some numbers can be written in two ways (1.8=1.7999999...) only consider the form without all the 9's.
Source: Slightly adapted from a post by Victor Wang on Facebook.
We want to prove that i
f x is a positive real number, then somewhere in the infinite sequence {x, 2x, 3x, ...} there is a number containing the digit 7.
We can easily prove something STRONGER, namely that somewhere in the infinite sequence {x, 2x, 3x, ...} there is a number STARTING with the digit 7.
In fact, we can prove something MUCH STRONGER, namely that in the infinite sequence {x, 2x, 3x, ...} there are an INFINITE NUMBER of numbers STARTING with ANY ARBITRARY SEQUENCE OF DIGITS.
PROOF:
Consider a base 10 slide rule. The set of all numbers beginning with an arbitrary set of digits occupy a continuous space on that slide rule. For instance, the set of all numbers beginning with 723 occupy the space [log 7.23, log 7.24). But the set of all positive integers have an infinite number of elements in every region on the slide rule, and so does the infinite sequence {x, 2x, 3x, ...}. Therefore there must be an infinite member of {x, 2x, 3x, ...} that begins with any arbitrary sequence of numbers.
PART II:
I have no idea how to find n.
Part I proof
