find a and b when 34a567b such that a and b is divisible by 3 and 11

Im sorry I analysed it wrng .My previous ans is right if the question is - find a and b when 34a567b such that the whole number 34a567b is perfectly divisible by 3 and 11.
but since u askd " find a and b when 34a567b such that a and b is divisible by 3 and 11?"-
Ans - a and b both can be 0 since 0 is divisible by all integers not only 3 and 11

( Also the qn should state perfectly divisible or else any no. can be divisible by any no. )
Tnq for the question . My frnd made me realize how wrng I was in jumping to conclusions without understanding the qn.

happy eegeh

I think its not possible .
Acc to divisibility rules for a no. to be divisible by 3 their sum of the digits should be a multiple / divisible by 3 i.e. 3+4+a+5+6+7+b=25+a+b/3 for that we need a+b to be either 2/5/8/11/14/17 so then itll be divisible by 3
and for a no to b divisible by 11 >>
3-4+a-5+6-7+b ( alternating signs from left ) = should b divisible by 11 . for that it needs a+b to be 7/18 which are different from what we require a and b to be when we need them to b divisible by 3 ( 2/5/8/11/14/17 )
check with ur answer

# New to this , my apologies if mistaken .