Find the smallest prime number that doesn't divide any 5-digit number whose digits are in strictly increasing order. Find the smallest prime number that doesn't divide any 5-digit number whose digits are in strictly increasing order.
I have posted an answer of my own below; any alternative solutions will also be greatly appreciated.
 A: *

*I first took the smallest possible 5-digit number with all digits in strictly increasing order: 12345. Now, it is obviously divisible by 3 & 5 (divisibility tests); this rules out 3 or 5 being the answer. 

*Then 12346 is obviously divisible by 2, hence 2 is also rules out
(as 12346 also has digits in strictly increasing order).

*Now, notice that remainder of 12345/7 is 4; this means that 7 is also rules out as it divides 12349.

*I then tried 11; none of them seemed to work. So I tried doing a generality test involving its divisibility rule.


For a number to be divisible by 11, the sum of the even digits is subtracted from the sum of the odd digits; the difference's absolute value is either 0 or divisible by 11.
So, I took a 5-digit number $abcde$ (Note: The letters have been concatenated).
$$S=a+c+e-b-d$$
$$S=a+(c-b)+(e-d)>a>0$$
$$S=e-(d-c)-(b-a)<e<10$$
(Second statement holds true because (c-b) and (e-d) are both positive, and Third statement is true because (b-a) and (d-c) are both positive as well).

Therefore: $0<S<10$,- making it impossible for the number to be divisible by 11 according to its divisibility rule; hence 11 is the smallest prime number satisfying the condition to not be able to divide any 5 digit number whose digits are in strictly increasing order.
