clearly 2,3,5 or7 are not the required primes (for 7,the counterexample is 12348) so let us check divisibility for 11 if $ a_1a_2a_3a_4a_5$ is divisible by 11 ,(where $a_1\ll a_2\ll a_3\ll a_4\ll a_5$) then $ (a_1-a_2)+(a_3-a_4)+a_5= 0 or 11$ and since $ a_5\le 9$ and terms in bracket are negative it should be equal to $ 0$ but when it is put in the form $ a_1+(-a_2+a_3)+(-a_4+a_5)=0 $ we see that it is impossible since terms in bracket are positive. So we conclude that no such number is divisible by 11.

I think when you write $(a_1-a_2) + (a_3-a_4) + a_5 = 0 \text{ or } 11$, you should also mention that it could equal $-11$ (although this is impossible by your second argument).

Note that $12346$ is even,$3$ and $5$ divide $12345$, and $7$ divides $12348$. Consider a $5$ digit number $n = abcde$ with $0 < a < b < c < d < e < 10$. Let $S$ = $(a+c+e)-(b+d)$. Then $S$=$a$+(c−b)+(e−d) $>a>0 $ and $S$ = e − (d − c) − (b − a) $< e$ ≤ $10$, so$ S $is not divisible by $11$ and hence n is not divisible by $11$. thus $11$ is the smallest prime that does not divide any five-digit number whose digits are in a strictly increasing order

Can we write like this in RMO? isn,t there any method that would give a definite answer

SS007 wrote: Can we write like this in RMO? isn,t there any method that would give a definite answer No