Problem 1753

Solution:

Such numbers exist. For a positive integer $m$ let $f(m)$ be the number of neckish numbers among the 102 consecutive numbers $m, m+1, m+2, \ldots, m+101$. Let $N$ be the least common multiple of the numbers $2,3,4, \ldots, 101$. Then $f\left(2^{N}\right) \geq 100$, since for all $b=2,3, \ldots, 101$ we have: $2^{N}+b=\left(2^{N / b}\right)^{b}+b$ is neckish. Thus there is also a smallest positive number $M$ with $f(M) \geq 100$. Clearly $M \leq 2^{N}$. On the other hand $M>1$ (and $M-1$ is a positive integer), because every neckish number is greater than 5, since for $a, b \geq 2$ we have: $a^{b}+b \geq 2^{b}+b \geq 2^{2}+2=6$, and thus among the numbers from 1 to 102 at most 97 are neckish. Now it is shown that $f(M)=100$: If $f(M)>100$ held, there would be among the numbers from $M$ to $M+101$ at least 101 neckish ones, hence among those from $M$ to $M+100$ at least 100 neckish ones, and likewise among those from $M-1$ to $M+100$. Then $f(M-1) \geq 100$ would hold, contradicting the minimality of $M$. Therefore $f(M)=100$. The numbers from $M$ to $M+101$ thus satisfy the condition.