Hopeless Numbers Beatriz Viterbo has called a positive integer which is not divisible by any of the ($2^n$, where $n$ is the number of its digits) numbers that result by introducing a plus or minus sign to the left of each of its digits a hopeless number.
Are there infinitely many hopeless numbers? Are there arbitrarily long strings of consecutive numbers all of which are hopeless?
 A: There cannot be more than $17$ consecutive hopeless numbers. This is because among any ten consecutive numbers, one of them must end in a $0$, and for any number, there is a signed digit sum between $0$ and $9$. (The latter fact is easily proved by induction.) If $N=10n$ and $0\le s\le9$ is one of its signed digit sums, then $N+1$ is hopeful if $s=0$, while $N+s-1$ is hopeful if $1\le s\le9$. That is, one of the next $8$ numbers has $1$ among its signed digit sums.
It would be of interest (to me, at least) to know what is the largest possible length of a consecutive string of hopeless numbers (with one or more explicit examples), as well as the largest length that occurs infinitely often. (The sequence $850,851,8500,8501,85000,85001,\ldots$ shows there are infinitely many consecutive pairs of hopeless numbers. Are there infinitely many consecutive triples?)
A: астон вілла олоф мэллбэрг answers the first question (there are infinitely many hopeless numbers).
For your second question, the answer is "no", there are not arbitrarily long sets of consecutive hopeless numbers. To see this, we will show that any positive integer has a resultant in the set $\{1,2,3,4,5,6,7,8,9,10,12,14,16,18\}$. It follows that any number divisible by $5040$ is not hopeless, and in particular there can never be $5040$ consecutive hopeless numbers.
To see this, insert signs $\pm$ one by one. For each sign except the sign before the last non-zero digit, choose the sign to minimise the absolute value of the sum so far. This means each such sum will be in the interval $[-9,+9]$. For the sign before the last non-zero digit, make sure the sum is non-zero, but if both possibilities are non-zero, choose the smaller absolute value. The only way you can be forced to choose a number with absolute value greater than $9$ is if the final non-zero digit is the same absolute value as the sum so far, in which case the absolute value of the final sum will be twice this (so even and at most $18$). Finally, if the sum is negative go back and swap all signs.
