# 3 digit number divisible by 3 and two digits repeated.....???

Aman Bansal
592 Points
11 years ago

DearShubham,

For the number to be divisible by , the number formed by the last two digits must be divisible by . With the digits  through , you can use the  last digits  and , and each of these can be combined with half of the digits to form a -digit number divisible by , so there are combinations for the last two digits. The other two digits can be chosen arbitrarily, so there are combinations for those, for a total of  different numbers

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums.

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

Peetala Kishore
33 Points
11 years ago

Hey

suppose 114

to test this = 1+1+4 =6 .

six is divisable by 3 so

the given number also divisable by 3.

there are 66 possiblities are there for your question.

they are = {114,117,225,228,330,336,339,441,448,552,558,660,669,771,774,882,885,990,993,996} there is also possibloe of repetitions  other are 300,600,900.