JOMO 6, Short 8

According to the divisibility test of 11, A number is divisible by 11 if the difference between $X$ and $Y$ where we define $X$ as sum of digits at odd numbered places from the right end of the number, and $Y$ as sum of digits at the even numbered places.

This divisibility test can be obtained from the following...

$10 \equiv -1\pmod{11} \implies 10^{2n} \equiv 1 \pmod{11}$ and $10^{2n+1} \equiv -1 \pmod{11}$ , and in the decimal representation, we will have all digits multiplied by a power of $10$ .

For a given set having digit sum 13, surely you can't split it into set of 2 equal digit sums (13 is odd ), so you will have to split the set in only 1 way, i.e. 12 and 1, so $11\mid 12-1$ .

This implies $1$ always has to be in the set.

NOTE- The solutions other than this posted till the moment either use incompletely assumed things or "trial and error" but i feel this solution gives complete proof of why 1 will be in the set....

the smallest no satisfying this con is 319 n v hv 3 tries.. lol... so... (simplest sol acc 2 me{Samiksha})

see at first sight we can say that the number is not a 2 digit number because the difference of digits would not be divisible by 11 and if difference of digits would be 0 then both digits same so there sum will be even but 13 is odd so number is not a 2 digit number. so it may be a 3 digit number or set may contain 3 digits. say a,b and c such that abc is divisible by 11 then a + c - b = 11 not 0 because then sum of digits would become even also not 22 or any further multiple of 11 bcoz a + b maximum val = 18. now a + c - b = 11 and a + b + c = 13 we get a + c = 12 and b = 1 we still can't declare one as the answer we will take up a 4 digit number abcd a+b+c+d = 13 and (a + c) - (b + d) = 11 so , a + c = 12 and b + d = 1 so here , b = 0 or 1 and d = 0 or 1 now we observe that we take any digit number the sum of the even placed digits = 1 . and odd placed digits as 12 so simply the digits which can have sum as 1 are 0 and 1 but in case of a 3 digit number 0 does not occur so 1 is the answer .\ NOTE : I HOPE THIS SOLUTION IS AN EXCEPTION TO ADITYA RAUT'S NOTE IN HIS SOLUTION

firstly try to get how many digits the collection has for the sum of digits =13 the collection should be minimum of two digits like 9+4=13 and the collection can have maximum of 5 digits as 0+1+2+3+4+5>13 so the collection can have maximum of 5 digits. now for the divisibility of 11 2 digits no cannot serve the result. so neglect it. for three digits no we can have sum of even digit =12 and odd digits =1 or viceversa(since the sum of digits need to be equal to 13). like wise for four and five digit no. so 1 is common in all that.