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.
For the first statement, we can prove the stronger statement that for any x, there exists a multiple of x that starts with a "7."
To do this, consider the interval [7 * 10^k, 8*10^k). If k is large enough that 10^k is greater than x, this ensures that at some point there will be a multiple of x that falls within the given interval.
As an example, if x = 99, then k must be greater than or equal to 2. So considering the interval 700 to 799, there must be at least one multiple of 99 included, since no two consecutive multiples of 99 can span from 699 to 800. Indeed 99*8 = 792.
Of course, this method doesn't get you the smallest such multiple of x that contains a 7 (although I suppose it does fix an upper bound on the value of n).
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Posted by H M
on 2022-08-12 11:06:44 |
Proof for 1st Part (not smallest)
